\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(i) }\quad 5x+7y+2=0,\qquad 4x+6y+3=0
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{The given equations are}
\displaystyle 5x+7y=-2
\displaystyle 4x+6y=-3
\displaystyle \text{The system can be written in matrix form as}
\displaystyle \begin{bmatrix}5&7\\4&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\-3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}5&7\\4&6\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}-2\\-3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}5&7\\4&6\end{vmatrix}=30-28=2\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ denote the cofactor of }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=6,\qquad C_{12}=-4,\qquad C_{21}=-7,\qquad C_{22}=5
\displaystyle \text{adj }A=\begin{bmatrix}6&-4\\-7&5\end{bmatrix}^{T}=\begin{bmatrix}6&-7\\-4&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A=\frac{1}{2}\begin{bmatrix}6&-7\\-4&5\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{2}\begin{bmatrix}6&-7\\-4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix}
\displaystyle =\frac{1}{2}\begin{bmatrix}-12+21\\8-15\end{bmatrix}=\frac{1}{2}\begin{bmatrix}9\\-7\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\-\frac{7}{2}\end{bmatrix}
\displaystyle \therefore x=\frac{9}{2}\text{ and }y=-\frac{7}{2}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(ii) }\quad 5x+2y=3,\qquad 3x+2y=5
\displaystyle \text{Answer:}
\displaystyle \text{(ii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}5&2\\3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}5&2\\3&2\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}5&2\\3&2\end{vmatrix}=10-6=4\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=2,\qquad C_{12}=-3,\qquad C_{21}=-2,\qquad C_{22}=5
\displaystyle \text{adj }A=\begin{bmatrix}2&-3\\-2&5\end{bmatrix}^{T}=\begin{bmatrix}2&-2\\-3&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-1\\4\end{bmatrix}
\displaystyle \therefore x=-1\text{ and }y=4.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(iii) }\quad 3x+4y-5=0,\qquad x-y+3=0
\displaystyle \text{Answer:}
\displaystyle \text{(iii)}
\displaystyle \text{The given equations are}
\displaystyle 3x+4y=5
\displaystyle x-y=-3
\displaystyle \text{The system can be written in matrix form as}
\displaystyle \begin{bmatrix}3&4\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\-3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&4\\1&-1\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}5\\-3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&4\\1&-1\end{vmatrix}=-3-4=-7\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=-1,\qquad C_{12}=-1,\qquad C_{21}=-4,\qquad C_{22}=3
\displaystyle \text{adj }A=\begin{bmatrix}-1&-1\\-4&3\end{bmatrix}^{T}=\begin{bmatrix}-1&-4\\-1&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-7}\begin{bmatrix}-1&-4\\-1&3\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{-7}\begin{bmatrix}-1&-4\\-1&3\end{bmatrix}\begin{bmatrix}5\\-3\end{bmatrix}
\displaystyle =\frac{1}{-7}\begin{bmatrix}-5+12\\-5-9\end{bmatrix}=\frac{1}{-7}\begin{bmatrix}7\\-14\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix}
\displaystyle \therefore x=-1\text{ and }y=2.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(iv) }\quad 3x+y=19,\qquad 3x-y=23
\displaystyle \text{Answer:}
\displaystyle \text{(iv)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&1\\3&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}19\\23\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&1\\3&-1\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}19\\23\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&1\\3&-1\end{vmatrix}=-3-3=-6\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=-1,\qquad C_{12}=-3,\qquad C_{21}=-1,\qquad C_{22}=3
\displaystyle \text{adj }A=\begin{bmatrix}-1&-3\\-1&3\end{bmatrix}^{T}=\begin{bmatrix}-1&-1\\-3&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-6}\begin{bmatrix}-1&-1\\-3&3\end{bmatrix}
\displaystyle X=A^{-1}B=\frac{1}{-6}\begin{bmatrix}-1&-1\\-3&3\end{bmatrix}\begin{bmatrix}19\\23\end{bmatrix}
\displaystyle =\frac{1}{-6}\begin{bmatrix}-19-23\\-57+69\end{bmatrix}=\frac{1}{-6}\begin{bmatrix}-42\\12\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-2\end{bmatrix}
\displaystyle \therefore x=7\text{ and }y=-2.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(v) }\quad 3x+7y=4,\qquad x+2y=-1
\displaystyle \text{Answer:}
\displaystyle \text{(v)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&7\\1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\-1\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&7\\1&2\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\-1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&7\\1&2\end{vmatrix}=6-7=-1\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=2,\qquad C_{12}=-1,\qquad C_{21}=-7,\qquad C_{22}=3
\displaystyle \text{adj }A=\begin{bmatrix}2&-1\\-7&3\end{bmatrix}^{T}=\begin{bmatrix}2&-7\\-1&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-1}\begin{bmatrix}2&-7\\-1&3\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\-1&3\end{bmatrix}\begin{bmatrix}4\\-1\end{bmatrix}
\displaystyle =\frac{1}{-1}\begin{bmatrix}8+7\\-4-3\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\-7\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-15\\7\end{bmatrix}
\displaystyle \therefore x=-15\text{ and }y=7.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(vi) }\quad 3x+y=7,\qquad 5x+3y=12
\displaystyle \text{Answer:}
\displaystyle \text{(vi)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&1\\5&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\12\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&1\\5&3\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}7\\12\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&1\\5&3\end{vmatrix}=9-5=4\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=3,\qquad C_{12}=-5,\qquad C_{21}=-1,\qquad C_{22}=3
\displaystyle \text{adj }A=\begin{bmatrix}3&-5\\-1&3\end{bmatrix}^{T}=\begin{bmatrix}3&-1\\-5&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{4}\begin{bmatrix}3&-1\\-5&3\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{4}\begin{bmatrix}3&-1\\-5&3\end{bmatrix}\begin{bmatrix}7\\12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}21-12\\-35+36\end{bmatrix}=\frac{1}{4}\begin{bmatrix}9\\1\end{bmatrix}
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\\frac{1}{4}\end{bmatrix}
\displaystyle \therefore x=\frac{9}{4}\text{ and }y=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(i) }\quad x+y-z=3,\qquad 2x+3y+z=10,\qquad 3x-y-7z=1
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&-1\\2&3&1\\3&-1&-7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\10\\1\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&-1\\2&3&1\\3&-1&-7\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}3\\10\\1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&-1\\2&3&1\\3&-1&-7\end{vmatrix}
\displaystyle =1(-21+1)-1(-14-3)-1(-2-9)
\displaystyle =-20+17+11=8\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}3&1\\-1&-7\end{vmatrix}=-20,\qquad C_{12}=-\begin{vmatrix}2&1\\3&-7\end{vmatrix}=17
\displaystyle C_{13}=\begin{vmatrix}2&3\\3&-1\end{vmatrix}=-11
\displaystyle C_{21}=-\begin{vmatrix}1&-1\\-1&-7\end{vmatrix}=8,\qquad C_{22}=\begin{vmatrix}1&-1\\3&-7\end{vmatrix}=-4
\displaystyle C_{23}=-\begin{vmatrix}1&1\\3&-1\end{vmatrix}=4
\displaystyle C_{31}=\begin{vmatrix}1&-1\\3&1\end{vmatrix}=4,\qquad C_{32}=-\begin{vmatrix}1&-1\\2&1\end{vmatrix}=-3
\displaystyle C_{33}=\begin{vmatrix}1&1\\2&3\end{vmatrix}=1
\displaystyle \text{adj }A=\begin{bmatrix}-20&17&-11\\8&-4&4\\4&-3&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-20&8&4\\17&-4&-3\\-11&4&1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{8}\begin{bmatrix}-20&8&4\\17&-4&-3\\-11&4&1\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{8}\begin{bmatrix}-20&8&4\\17&-4&-3\\-11&4&1\end{bmatrix}\begin{bmatrix}3\\10\\1\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}-60+80+4\\51-40-3\\-33+40+1\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}24\\8\\8\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}
\displaystyle \therefore x=3,\ y=1\text{ and }z=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(ii) }\quad x+y+z=3,\qquad 2x-y+z=-1
\displaystyle 2x+y-3z=-9.\hfill \text{[CBSE 2004, 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{(ii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\2&-1&1\\2&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\-1\\-9\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\2&-1&1\\2&1&-3\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}3\\-1\\-9\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\2&-1&1\\2&1&-3\end{vmatrix}
\displaystyle =1(3-1)-1(-6-2)+1(2+2)
\displaystyle =2+8+4=14\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-1&1\\1&-3\end{vmatrix}=2,\qquad C_{12}=-\begin{vmatrix}2&1\\2&-3\end{vmatrix}=8
\displaystyle C_{13}=\begin{vmatrix}2&-1\\2&1\end{vmatrix}=4
\displaystyle C_{21}=-\begin{vmatrix}1&1\\1&-3\end{vmatrix}=4,\qquad C_{22}=\begin{vmatrix}1&1\\2&-3\end{vmatrix}=-5
\displaystyle C_{23}=-\begin{vmatrix}1&1\\2&1\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}1&1\\-1&1\end{vmatrix}=2,\qquad C_{32}=-\begin{vmatrix}1&1\\2&1\end{vmatrix}=1
\displaystyle C_{33}=\begin{vmatrix}1&1\\2&-1\end{vmatrix}=-3
\displaystyle \text{adj }A=\begin{bmatrix}2&8&4\\4&-5&1\\2&1&-3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}2&4&2\\8&-5&1\\4&1&-3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{14}\begin{bmatrix}2&4&2\\8&-5&1\\4&1&-3\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{14}\begin{bmatrix}2&4&2\\8&-5&1\\4&1&-3\end{bmatrix}\begin{bmatrix}3\\-1\\-9\end{bmatrix}
\displaystyle =\frac{1}{14}\begin{bmatrix}6-4-18\\24+5-9\\12-1+27\end{bmatrix}
\displaystyle =\frac{1}{14}\begin{bmatrix}-16\\20\\38\end{bmatrix}=\begin{bmatrix}-\frac{8}{7}\\\frac{10}{7}\\\frac{19}{7}\end{bmatrix}
\displaystyle \therefore x=-\frac{8}{7},\ y=\frac{10}{7}\text{ and }z=\frac{19}{7}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(iii) }\quad 6x-12y+25z=4,\qquad 4x+15y-20z=3
\displaystyle 2x+18y+15z=10
\displaystyle \text{Answer:}
\displaystyle \text{(iii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}6&-12&25\\4&15&-20\\2&18&15\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\3\\10\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}6&-12&25\\4&15&-20\\2&18&15\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\3\\10\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}6&-12&25\\4&15&-20\\2&18&15\end{vmatrix}
\displaystyle =6(225+360)+12(60+40)+25(72-30)
\displaystyle =3510+1200+1050=5760\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}15&-20\\18&15\end{vmatrix}=585,\qquad C_{12}=-\begin{vmatrix}4&-20\\2&15\end{vmatrix}=-100
\displaystyle C_{13}=\begin{vmatrix}4&15\\2&18\end{vmatrix}=42
\displaystyle C_{21}=-\begin{vmatrix}-12&25\\18&15\end{vmatrix}=630,\qquad C_{22}=\begin{vmatrix}6&25\\2&15\end{vmatrix}=40
\displaystyle C_{23}=-\begin{vmatrix}6&-12\\2&18\end{vmatrix}=-132
\displaystyle C_{31}=\begin{vmatrix}-12&25\\15&-20\end{vmatrix}=-135,\qquad C_{32}=-\begin{vmatrix}6&25\\4&-20\end{vmatrix}=220
\displaystyle C_{33}=\begin{vmatrix}6&-12\\4&15\end{vmatrix}=138
\displaystyle \text{adj }A=\begin{bmatrix}585&-100&42\\630&40&-132\\-135&220&138\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}585&630&-135\\-100&40&220\\42&-132&138\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{5760}\begin{bmatrix}585&630&-135\\-100&40&220\\42&-132&138\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{5760}\begin{bmatrix}585&630&-135\\-100&40&220\\42&-132&138\end{bmatrix}\begin{bmatrix}4\\3\\10\end{bmatrix}
\displaystyle =\frac{1}{5760}\begin{bmatrix}2340+1890-1350\\-400+120+2200\\168-396+1380\end{bmatrix}
\displaystyle =\frac{1}{5760}\begin{bmatrix}2880\\1920\\1152\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\\frac{1}{3}\\\frac{1}{5}\end{bmatrix}
\displaystyle \therefore x=\frac{1}{2},\ y=\frac{1}{3}\text{ and }z=\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(iv) }\quad 3x+4y+7z=14,\qquad 2x-y+3z=4
\displaystyle x+2y-3z=0
\displaystyle \text{Answer:}
\displaystyle \text{(iv)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&4&7\\2&-1&3\\1&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}14\\4\\0\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&4&7\\2&-1&3\\1&2&-3\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}14\\4\\0\end{bmatrix}
\displaystyle |A|=3\begin{vmatrix}-1&3\\2&-3\end{vmatrix}-4\begin{vmatrix}2&3\\1&-3\end{vmatrix}+7\begin{vmatrix}2&-1\\1&2\end{vmatrix}
\displaystyle =3(3-6)-4(-6-3)+7(4+1)
\displaystyle =-9+36+35=62\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-1&3\\2&-3\end{vmatrix}=-3,\qquad C_{12}=-\begin{vmatrix}2&3\\1&-3\end{vmatrix}=9
\displaystyle C_{13}=\begin{vmatrix}2&-1\\1&2\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}4&7\\2&-3\end{vmatrix}=26,\qquad C_{22}=\begin{vmatrix}3&7\\1&-3\end{vmatrix}=-16
\displaystyle C_{23}=-\begin{vmatrix}3&4\\1&2\end{vmatrix}=-2
\displaystyle C_{31}=\begin{vmatrix}4&7\\-1&3\end{vmatrix}=19,\qquad C_{32}=-\begin{vmatrix}3&7\\2&3\end{vmatrix}=5
\displaystyle C_{33}=\begin{vmatrix}3&4\\2&-1\end{vmatrix}=-11
\displaystyle \text{adj }A=\begin{bmatrix}-3&9&5\\26&-16&-2\\19&5&-11\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{62}\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{62}\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}\begin{bmatrix}14\\4\\0\end{bmatrix}
\displaystyle =\frac{1}{62}\begin{bmatrix}-42+104+0\\126-64+0\\70-8+0\end{bmatrix}
\displaystyle =\frac{1}{62}\begin{bmatrix}62\\62\\62\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}
\displaystyle \therefore x=1,\ y=1\text{ and }z=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(v) }\quad \frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10,\qquad \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10
\displaystyle \frac{3}{x}-\frac{1}{y}+\frac{2}{z}=13.\hfill \text{[CBSE 2005, 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{(v)}
\displaystyle \text{Let }a=\frac{1}{x},\ b=\frac{1}{y}\text{ and }c=\frac{1}{z}.
\displaystyle \text{Then the given system becomes}
\displaystyle 2a-3b+3c=10,\qquad a+b+c=10,\qquad 3a-b+2c=13.
\displaystyle \text{The system can be written in matrix form as}
\displaystyle \begin{bmatrix}2&-3&3\\1&1&1\\3&-1&2\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}10\\10\\13\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&-3&3\\1&1&1\\3&-1&2\end{bmatrix},\ X=\begin{bmatrix}a\\b\\c\end{bmatrix}\text{ and }B=\begin{bmatrix}10\\10\\13\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&-3&3\\1&1&1\\3&-1&2\end{vmatrix}
\displaystyle =2(2+1)+3(2-3)+3(-1-3)
\displaystyle =6-3-12=-9\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}1&1\\-1&2\end{vmatrix}=3,\qquad C_{12}=-\begin{vmatrix}1&1\\3&2\end{vmatrix}=1,\qquad C_{13}=\begin{vmatrix}1&1\\3&-1\end{vmatrix}=-4
\displaystyle C_{21}=-\begin{vmatrix}-3&3\\-1&2\end{vmatrix}=3,\qquad C_{22}=\begin{vmatrix}2&3\\3&2\end{vmatrix}=-5,\qquad C_{23}=-\begin{vmatrix}2&-3\\3&-1\end{vmatrix}=-7
\displaystyle C_{31}=\begin{vmatrix}-3&3\\1&1\end{vmatrix}=-6,\qquad C_{32}=-\begin{vmatrix}2&3\\1&1\end{vmatrix}=1,\qquad C_{33}=\begin{vmatrix}2&-3\\1&1\end{vmatrix}=5
\displaystyle \text{adj }A=\begin{bmatrix}3&1&-4\\3&-5&-7\\-6&1&5\end{bmatrix}^{T}=\begin{bmatrix}3&3&-6\\1&-5&1\\-4&-7&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-9}\begin{bmatrix}3&3&-6\\1&-5&1\\-4&-7&5\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}a\\b\\c\end{bmatrix}=\frac{1}{-9}\begin{bmatrix}3&3&-6\\1&-5&1\\-4&-7&5\end{bmatrix}\begin{bmatrix}10\\10\\13\end{bmatrix}
\displaystyle =\frac{1}{-9}\begin{bmatrix}30+30-78\\10-50+13\\-40-70+65\end{bmatrix}=\frac{1}{-9}\begin{bmatrix}-18\\-27\\-45\end{bmatrix}
\displaystyle \therefore a=2,\ b=3,\ c=5
\displaystyle \therefore x=\frac{1}{a}=\frac{1}{2},\quad y=\frac{1}{b}=\frac{1}{3},\quad z=\frac{1}{c}=\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(vi) }\quad 5x+3y+z=16,\qquad 2x+y+3z=19
\displaystyle x+2y+4z=25
\displaystyle \text{Answer:}
\displaystyle \text{(vi)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}16\\19\\25\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}16\\19\\25\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}5&3&1\\2&1&3\\1&2&4\end{vmatrix}
\displaystyle =5(4-6)-3(8-3)+(4-1)
\displaystyle =-10-15+3=-22\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}1&3\\2&4\end{vmatrix}=-2,\qquad C_{12}=-\begin{vmatrix}2&3\\1&4\end{vmatrix}=-5,\qquad C_{13}=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3
\displaystyle C_{21}=-\begin{vmatrix}3&1\\2&4\end{vmatrix}=-10,\qquad C_{22}=\begin{vmatrix}5&1\\1&4\end{vmatrix}=19,\qquad C_{23}=-\begin{vmatrix}5&3\\1&2\end{vmatrix}=-7
\displaystyle C_{31}=\begin{vmatrix}3&1\\1&3\end{vmatrix}=8,\qquad C_{32}=-\begin{vmatrix}5&1\\2&3\end{vmatrix}=-13,\qquad C_{33}=\begin{vmatrix}5&3\\2&1\end{vmatrix}=-1
\displaystyle \text{adj }A=\begin{bmatrix}-2&-5&3\\-10&19&-7\\8&-13&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}\begin{bmatrix}16\\19\\25\end{bmatrix}
\displaystyle =\frac{1}{-22}\begin{bmatrix}-32-190+200\\-80+361-325\\48-133-25\end{bmatrix}
\displaystyle =\frac{1}{-22}\begin{bmatrix}-22\\-44\\-110\end{bmatrix}=\begin{bmatrix}1\\2\\5\end{bmatrix}
\displaystyle \therefore x=1,\ y=2\text{ and }z=5.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(vii) }\quad 3x+4y+2z=8,\qquad 2y-3z=3
\displaystyle x-2y+6z=-2
\displaystyle \text{Answer:}
\displaystyle \text{(vii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&4&2\\0&2&-3\\1&-2&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8\\3\\-2\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&4&2\\0&2&-3\\1&-2&6\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}8\\3\\-2\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&4&2\\0&2&-3\\1&-2&6\end{vmatrix}
\displaystyle =3(12-6)-4(0+3)+2(0-2)
\displaystyle =18-12-4=2\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&-3\\-2&6\end{vmatrix}=6,\qquad C_{12}=-\begin{vmatrix}0&-3\\1&6\end{vmatrix}=-3,\qquad C_{13}=\begin{vmatrix}0&2\\1&-2\end{vmatrix}=-2
\displaystyle C_{21}=-\begin{vmatrix}4&2\\-2&6\end{vmatrix}=-28,\qquad C_{22}=\begin{vmatrix}3&2\\1&6\end{vmatrix}=16,\qquad C_{23}=-\begin{vmatrix}3&4\\1&-2\end{vmatrix}=10
\displaystyle C_{31}=\begin{vmatrix}4&2\\2&-3\end{vmatrix}=-16,\qquad C_{32}=-\begin{vmatrix}3&2\\0&-3\end{vmatrix}=9,\qquad C_{33}=\begin{vmatrix}3&4\\0&2\end{vmatrix}=6
\displaystyle \text{adj }A=\begin{bmatrix}6&-3&-2\\-28&16&10\\-16&9&6\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}6&-28&-16\\-3&16&9\\-2&10&6\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{2}\begin{bmatrix}6&-28&-16\\-3&16&9\\-2&10&6\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{2}\begin{bmatrix}6&-28&-16\\-3&16&9\\-2&10&6\end{bmatrix}\begin{bmatrix}8\\3\\-2\end{bmatrix}
\displaystyle =\frac{1}{2}\begin{bmatrix}48-84+32\\-24+48-18\\-16+30-12\end{bmatrix}
\displaystyle =\frac{1}{2}\begin{bmatrix}-4\\6\\2\end{bmatrix}=\begin{bmatrix}-2\\3\\1\end{bmatrix}
\displaystyle \therefore x=-2,\ y=3\text{ and }z=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(viii) }\quad 2x+y+z=2,\qquad x+3y-z=5
\displaystyle 3x+y-2z=6.\hfill \text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{(viii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&1&1\\1&3&-1\\3&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\5\\6\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&1&1\\1&3&-1\\3&1&-2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\5\\6\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&1&1\\1&3&-1\\3&1&-2\end{vmatrix}
\displaystyle =2(-6+1)-(-2+3)+(1-9)
\displaystyle =-10-1-8=-19\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}3&-1\\1&-2\end{vmatrix}=-5,\qquad C_{12}=-\begin{vmatrix}1&-1\\3&-2\end{vmatrix}=-1,\qquad C_{13}=\begin{vmatrix}1&3\\3&1\end{vmatrix}=-8
\displaystyle C_{21}=-\begin{vmatrix}1&1\\1&-2\end{vmatrix}=3,\qquad C_{22}=\begin{vmatrix}2&1\\3&-2\end{vmatrix}=-7,\qquad C_{23}=-\begin{vmatrix}2&1\\3&1\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}1&1\\3&-1\end{vmatrix}=-4,\qquad C_{32}=-\begin{vmatrix}2&1\\1&-1\end{vmatrix}=3,\qquad C_{33}=\begin{vmatrix}2&1\\1&3\end{vmatrix}=5
\displaystyle \text{adj }A=\begin{bmatrix}-5&-1&-8\\3&-7&1\\-4&3&5\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-5&3&-4\\-1&-7&3\\-8&1&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-19}\begin{bmatrix}-5&3&-4\\-1&-7&3\\-8&1&5\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-19}\begin{bmatrix}-5&3&-4\\-1&-7&3\\-8&1&5\end{bmatrix}\begin{bmatrix}2\\5\\6\end{bmatrix}
\displaystyle =\frac{1}{-19}\begin{bmatrix}-10+15-24\\-2-35+18\\-16+5+30\end{bmatrix}
\displaystyle =\frac{1}{-19}\begin{bmatrix}-19\\-19\\19\end{bmatrix}=\begin{bmatrix}1\\1\\-1\end{bmatrix}
\displaystyle \therefore x=1,\ y=1\text{ and }z=-1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(ix) }\quad 2x+6y=2,\qquad 3x-z=-8
\displaystyle 2x-y+z=-3.\hfill \text{[CBSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{(ix)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&6&0\\3&0&-1\\2&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-8\\-3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&6&0\\3&0&-1\\2&-1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\-8\\-3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&6&0\\3&0&-1\\2&-1&1\end{vmatrix}
\displaystyle =2(0-1)-6(3+2)+0(-3-0)
\displaystyle =-2-30=-32\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}0&-1\\-1&1\end{vmatrix}=-1,\qquad C_{12}=-\begin{vmatrix}3&-1\\2&1\end{vmatrix}=-5,\qquad C_{13}=\begin{vmatrix}3&0\\2&-1\end{vmatrix}=-3
\displaystyle C_{21}=-\begin{vmatrix}6&0\\-1&1\end{vmatrix}=-6,\qquad C_{22}=\begin{vmatrix}2&0\\2&1\end{vmatrix}=2,\qquad C_{23}=-\begin{vmatrix}2&6\\2&-1\end{vmatrix}=14
\displaystyle C_{31}=\begin{vmatrix}6&0\\0&-1\end{vmatrix}=-6,\qquad C_{32}=-\begin{vmatrix}2&0\\3&-1\end{vmatrix}=2,\qquad C_{33}=\begin{vmatrix}2&6\\3&0\end{vmatrix}=-18
\displaystyle \text{adj }A=\begin{bmatrix}-1&-5&-3\\-6&2&14\\-6&2&-18\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&-6&-6\\-5&2&2\\-3&14&-18\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-32}\begin{bmatrix}-1&-6&-6\\-5&2&2\\-3&14&-18\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-32}\begin{bmatrix}-1&-6&-6\\-5&2&2\\-3&14&-18\end{bmatrix}\begin{bmatrix}2\\-8\\-3\end{bmatrix}
\displaystyle =\frac{1}{-32}\begin{bmatrix}-2+48+18\\-10-16-6\\-6-112+54\end{bmatrix}
\displaystyle =\frac{1}{-32}\begin{bmatrix}64\\-32\\-64\end{bmatrix}=\begin{bmatrix}-2\\1\\2\end{bmatrix}
\displaystyle \therefore x=-2,\ y=1\text{ and }z=2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(x) }\quad x-y+z=2,\qquad 2x-y=0
\displaystyle 2y-z=1.\hfill \text{[CBSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{(x)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-1&1\\2&-1&0\\0&2&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\0\\1\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&-1&1\\2&-1&0\\0&2&-1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\0\\1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&-1&1\\2&-1&0\\0&2&-1\end{vmatrix}
\displaystyle =1(1-0)+1(-2-0)+1(4-0)
\displaystyle =1-2+4=3\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-1&0\\2&-1\end{vmatrix}=1,\qquad C_{12}=-\begin{vmatrix}2&0\\0&-1\end{vmatrix}=2,\qquad C_{13}=\begin{vmatrix}2&-1\\0&2\end{vmatrix}=4
\displaystyle C_{21}=-\begin{vmatrix}-1&1\\2&-1\end{vmatrix}=1,\qquad C_{22}=\begin{vmatrix}1&1\\0&-1\end{vmatrix}=-1,\qquad C_{23}=-\begin{vmatrix}1&-1\\0&2\end{vmatrix}=-2
\displaystyle C_{31}=\begin{vmatrix}-1&1\\-1&0\end{vmatrix}=1,\qquad C_{32}=-\begin{vmatrix}1&1\\2&0\end{vmatrix}=2,\qquad C_{33}=\begin{vmatrix}1&-1\\2&-1\end{vmatrix}=1
\displaystyle \text{adj }A=\begin{bmatrix}1&2&4\\1&-1&-2\\1&2&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}1&1&1\\2&-1&2\\4&-2&1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{3}\begin{bmatrix}1&1&1\\2&-1&2\\4&-2&1\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{3}\begin{bmatrix}1&1&1\\2&-1&2\\4&-2&1\end{bmatrix}\begin{bmatrix}2\\0\\1\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}2+0+1\\4+0+2\\8+0+1\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}3\\6\\9\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \therefore x=1,\ y=2\text{ and }z=3.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(xi) }\quad 8x+4y+3z=18,\qquad 2x+y+z=5
\displaystyle x+2y+z=5.\hfill \text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{(xi)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}8&4&3\\2&1&1\\1&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}18\\5\\5\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}8&4&3\\2&1&1\\1&2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}18\\5\\5\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}8&4&3\\2&1&1\\1&2&1\end{vmatrix}
\displaystyle =8(1-2)-4(2-1)+3(4-1)
\displaystyle =-8-4+9=-3\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,\qquad C_{12}=-\begin{vmatrix}2&1\\1&1\end{vmatrix}=-1,\qquad C_{13}=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3
\displaystyle C_{21}=-\begin{vmatrix}4&3\\2&1\end{vmatrix}=2,\qquad C_{22}=\begin{vmatrix}8&3\\1&1\end{vmatrix}=5,\qquad C_{23}=-\begin{vmatrix}8&4\\1&2\end{vmatrix}=-12
\displaystyle C_{31}=\begin{vmatrix}4&3\\1&1\end{vmatrix}=1,\qquad C_{32}=-\begin{vmatrix}8&3\\2&1\end{vmatrix}=-2,\qquad C_{33}=\begin{vmatrix}8&4\\2&1\end{vmatrix}=0
\displaystyle \text{adj }A=\begin{bmatrix}-1&-1&3\\2&5&-12\\1&-2&0\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&2&1\\-1&5&-2\\3&-12&0\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{-3}\begin{bmatrix}-1&2&1\\-1&5&-2\\3&-12&0\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-1&2&1\\-1&5&-2\\3&-12&0\end{bmatrix}\begin{bmatrix}18\\5\\5\end{bmatrix}
\displaystyle =\frac{1}{-3}\begin{bmatrix}-18+10+5\\-18+25-10\\54-60+0\end{bmatrix}
\displaystyle =\frac{1}{-3}\begin{bmatrix}-3\\-3\\-6\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}
\displaystyle \therefore x=1,\ y=1\text{ and }z=2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(xii) }\quad x+y+z=6,\qquad x+2z=7
\displaystyle 3x+y+z=12.\hfill \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{(xii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\1&0&2\\3&1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&0&2\\3&1&1\end{vmatrix}
\displaystyle =1(0-2)-1(1-6)+1(1-0)
\displaystyle =-2+5+1=4\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}0&2\\1&1\end{vmatrix}=-2,\qquad C_{12}=-\begin{vmatrix}1&2\\3&1\end{vmatrix}=5,\qquad C_{13}=\begin{vmatrix}1&0\\3&1\end{vmatrix}=1
\displaystyle C_{21}=-\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,\qquad C_{22}=\begin{vmatrix}1&1\\3&1\end{vmatrix}=-2,\qquad C_{23}=-\begin{vmatrix}1&1\\3&1\end{vmatrix}=2
\displaystyle C_{31}=\begin{vmatrix}1&1\\0&2\end{vmatrix}=2,\qquad C_{32}=-\begin{vmatrix}1&1\\1&2\end{vmatrix}=-1,\qquad C_{33}=\begin{vmatrix}1&1\\1&0\end{vmatrix}=-1
\displaystyle \text{adj }A=\begin{bmatrix}-2&5&1\\0&-2&2\\2&-1&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-2&0&2\\5&-2&-1\\1&2&-1\end{bmatrix}\begin{bmatrix}6\\7\\12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}-12+0+24\\30-14-12\\6+14-12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}12\\4\\8\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}
\displaystyle \therefore x=3,\ y=1\text{ and }z=2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(xiii) }\quad \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\qquad \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1
\displaystyle \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2,\quad x,y,z\ne0.\hfill \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{(xiii)}
\displaystyle \text{Let }a=\frac{1}{x},\quad b=\frac{1}{y}\quad\text{and}\quad c=\frac{1}{z}.
\displaystyle \text{The given system of equations becomes}
\displaystyle 2a+3b+10c=4
\displaystyle 4a-6b+5c=1
\displaystyle 6a+9b-20c=2
\displaystyle \text{These equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\ X=\begin{bmatrix}a\\b\\c\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3&10\\4&-6&5\\6&9&-20\end{vmatrix}
\displaystyle =2(120-45)-3(-80-30)+10(36+36)
\displaystyle =150+330+720=1200\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-6&5\\9&-20\end{vmatrix}=75,\qquad C_{12}=-\begin{vmatrix}4&5\\6&-20\end{vmatrix}=110,\qquad C_{13}=\begin{vmatrix}4&-6\\6&9\end{vmatrix}=72
\displaystyle C_{21}=-\begin{vmatrix}3&10\\9&-20\end{vmatrix}=150,\qquad C_{22}=\begin{vmatrix}2&10\\6&-20\end{vmatrix}=-100,\qquad C_{23}=-\begin{vmatrix}2&3\\6&9\end{vmatrix}=0
\displaystyle C_{31}=\begin{vmatrix}3&10\\-6&5\end{vmatrix}=75,\qquad C_{32}=-\begin{vmatrix}2&10\\4&5\end{vmatrix}=30,\qquad C_{33}=\begin{vmatrix}2&3\\4&-6\end{vmatrix}=-24
\displaystyle \text{adj }A=\begin{bmatrix}75&110&72\\150&-100&0\\75&30&-24\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}a\\b\\c\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}
\displaystyle =\frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix}
\displaystyle =\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}=\begin{bmatrix}\frac12\\\frac13\\\frac15\end{bmatrix}
\displaystyle \therefore a=\frac12,\quad b=\frac13\quad\text{and}\quad c=\frac15.
\displaystyle \text{Since }a=\frac1x,\ b=\frac1y\text{ and }c=\frac1z,
\displaystyle x=\frac1a=2,\qquad y=\frac1b=3,\qquad z=\frac1c=5.
\displaystyle \therefore x=2,\ y=3\text{ and }z=5.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the matrix method:}
\displaystyle \text{(xiv) }\quad x-y+2z=7,\qquad 3x+4y-5z=-5
\displaystyle 2x-y+3z=12.\hfill \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{(xiv)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-5\\12\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}7\\-5\\12\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{vmatrix}
\displaystyle =1(12-5)+1(9+10)+2(-3-8)
\displaystyle =7+19-22=4\ne0
\displaystyle \therefore \text{The system has a unique solution given by }X=A^{-1}B.
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}4&-5\\-1&3\end{vmatrix}=7,\qquad C_{12}=-\begin{vmatrix}3&-5\\2&3\end{vmatrix}=-19,\qquad C_{13}=\begin{vmatrix}3&4\\2&-1\end{vmatrix}=-11
\displaystyle C_{21}=-\begin{vmatrix}-1&2\\-1&3\end{vmatrix}=1,\qquad C_{22}=\begin{vmatrix}1&2\\2&3\end{vmatrix}=-1,\qquad C_{23}=-\begin{vmatrix}1&-1\\2&-1\end{vmatrix}=-1
\displaystyle C_{31}=\begin{vmatrix}-1&2\\4&-5\end{vmatrix}=-3,\qquad C_{32}=-\begin{vmatrix}1&2\\3&-5\end{vmatrix}=11,\qquad C_{33}=\begin{vmatrix}1&-1\\3&4\end{vmatrix}=7
\displaystyle \text{adj }A=\begin{bmatrix}7&-19&-11\\1&-1&-1\\-3&11&7\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{ adj }A=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}
\displaystyle =\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}
\displaystyle \therefore x=2,\ y=1\text{ and }z=3.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(i) }\quad 6x+4y=2,\qquad 9x+6y=3
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle 6x+4y=2\qquad\ldots(1)
\displaystyle 9x+6y=3\qquad\ldots(2)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}6&4\\9&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}6&4\\9&6\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}6&4\\9&6\end{vmatrix}=36-36=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=6,\qquad C_{12}=-9,\qquad C_{21}=-4,\qquad C_{22}=6
\displaystyle \text{adj }A=\begin{bmatrix}6&-9\\-4&6\end{bmatrix}^{T}=\begin{bmatrix}6&-4\\-9&6\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}6&-4\\-9&6\end{bmatrix}\begin{bmatrix}2\\3\end{bmatrix}
\displaystyle =\begin{bmatrix}12-12\\-18+18\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }y=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }y=k\text{ in equation }(1),\text{ we get}
\displaystyle 6x+4k=2
\displaystyle \Rightarrow 6x=2-4k
\displaystyle \Rightarrow x=\frac{2-4k}{6}=\frac{1-2k}{3}
\displaystyle \therefore x=\frac{1-2k}{3}\text{ and }y=k,\quad k\in\mathbb{R}.
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(ii) }\quad 2x+3y=5,\qquad 6x+9y=15
\displaystyle \text{Answer:}
\displaystyle \text{(ii)}
\displaystyle 2x+3y=5\qquad\ldots(1)
\displaystyle 6x+9y=15\qquad\ldots(2)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&3\\6&9\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\15\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&3\\6&9\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}5\\15\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3\\6&9\end{vmatrix}=18-18=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=9,\qquad C_{12}=-6,\qquad C_{21}=-3,\qquad C_{22}=2
\displaystyle \text{adj }A=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^{T}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}
\displaystyle =\begin{bmatrix}45-45\\-30+30\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }y=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }y=k\text{ in equation }(1),\text{ we get}
\displaystyle 2x+3k=5
\displaystyle \Rightarrow 2x=5-3k
\displaystyle \Rightarrow x=\frac{5-3k}{2}
\displaystyle \therefore x=\frac{5-3k}{2}\text{ and }y=k,\quad k\in\mathbb{R}.
\displaystyle \text{These values of }x\text{ and }y\text{ also satisfy equation }(2).
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(iii) }\quad 5x+3y+7z=4,\qquad 3x+26y+2z=9, \qquad   7x+2y+10z=5
\displaystyle \text{Answer:}
\displaystyle \text{(iii)}
\displaystyle 5x+3y+7z=4\qquad\ldots(1)
\displaystyle 3x+26y+2z=9\qquad\ldots(2)
\displaystyle 7x+2y+10z=5\qquad\ldots(3)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\9\\5\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}5&3&7\\3&26&2\\7&2&10\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}4\\9\\5\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}5&3&7\\3&26&2\\7&2&10\end{vmatrix}
\displaystyle =5(260-4)-3(30-14)+7(6-182)
\displaystyle =1280-48-1232=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}26&2\\2&10\end{vmatrix}=256,\qquad C_{12}=-\begin{vmatrix}3&2\\7&10\end{vmatrix}=-16,\qquad C_{13}=\begin{vmatrix}3&26\\7&2\end{vmatrix}=-176
\displaystyle C_{21}=-\begin{vmatrix}3&7\\2&10\end{vmatrix}=-16,\qquad C_{22}=\begin{vmatrix}5&7\\7&10\end{vmatrix}=1,\qquad C_{23}=-\begin{vmatrix}5&3\\7&2\end{vmatrix}=11
\displaystyle C_{31}=\begin{vmatrix}3&7\\26&2\end{vmatrix}=-176,\qquad C_{32}=-\begin{vmatrix}5&7\\3&2\end{vmatrix}=11,\qquad C_{33}=\begin{vmatrix}5&3\\3&26\end{vmatrix}=121
\displaystyle \text{adj }A=\begin{bmatrix}256&-16&-176\\-16&1&11\\-176&11&121\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}256&-16&-176\\-16&1&11\\-176&11&121\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}256&-16&-176\\-16&1&11\\-176&11&121\end{bmatrix}\begin{bmatrix}4\\9\\5\end{bmatrix}
\displaystyle =\begin{bmatrix}1024-144-880\\-64+9+55\\-704+99+605\end{bmatrix}
\displaystyle =\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }z=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }z=k\text{ in equations }(1)\text{ and }(2),\text{ we get}
\displaystyle 5x+3y=4-7k
\displaystyle 3x+26y=9-2k
\displaystyle \begin{bmatrix}5&3\\3&26\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4-7k\\9-2k\end{bmatrix}
\displaystyle A_1X_1=B_1,\text{ where }A_1=\begin{bmatrix}5&3\\3&26\end{bmatrix},\ X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B_1=\begin{bmatrix}4-7k\\9-2k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}5&3\\3&26\end{vmatrix}=130-9=121\ne0
\displaystyle \text{adj }A_1=\begin{bmatrix}26&-3\\-3&5\end{bmatrix}
\displaystyle A_1^{-1}=\frac{1}{|A_1|}\text{ adj }A_1=\frac{1}{121}\begin{bmatrix}26&-3\\-3&5\end{bmatrix}
\displaystyle X_1=A_1^{-1}B_1
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{121}\begin{bmatrix}26&-3\\-3&5\end{bmatrix}\begin{bmatrix}4-7k\\9-2k\end{bmatrix}
\displaystyle =\frac{1}{121}\begin{bmatrix}104-182k-27+6k\\-12+21k+45-10k\end{bmatrix}
\displaystyle =\frac{1}{121}\begin{bmatrix}77-176k\\33+11k\end{bmatrix}
\displaystyle =\begin{bmatrix}\dfrac{7-16k}{11}\\[4pt]\dfrac{3+k}{11}\end{bmatrix}
\displaystyle \therefore x=\frac{7-16k}{11},\qquad y=\frac{3+k}{11}\qquad\text{and}\qquad z=k,\quad k\in\mathbb{R}.
\displaystyle \text{These values of }x,\ y\text{ and }z\text{ also satisfy equation }(3).
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(iv) }\quad x-y+z=3,\qquad 2x+y-z=2  , \qquad  -x-2y+2z=1
\displaystyle \text{Answer:}
\displaystyle \text{(iv)}
\displaystyle x-y+z=3\qquad\ldots(1)
\displaystyle 2x+y-z=2\qquad\ldots(2)
\displaystyle -x-2y+2z=1\qquad\ldots(3)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\2\\1\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}3\\2\\1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}
\displaystyle =1(2-2)+1(4-1)+1(-4+1)
\displaystyle =0+3-3=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}1&-1\\-2&2\end{vmatrix}=0,\qquad C_{12}=-\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=-3,\qquad C_{13}=\begin{vmatrix}2&1\\-1&-2\end{vmatrix}=-3
\displaystyle C_{21}=-\begin{vmatrix}-1&1\\-2&2\end{vmatrix}=0,\qquad C_{22}=\begin{vmatrix}1&1\\-1&2\end{vmatrix}=3,\qquad C_{23}=-\begin{vmatrix}1&-1\\-1&-2\end{vmatrix}=3
\displaystyle C_{31}=\begin{vmatrix}-1&1\\1&-1\end{vmatrix}=0,\qquad C_{32}=-\begin{vmatrix}1&1\\2&-1\end{vmatrix}=3,\qquad C_{33}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=3
\displaystyle \text{adj }A=\begin{bmatrix}0&-3&-3\\0&3&3\\0&3&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}0&0&0\\-3&3&3\\-3&3&3\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}0&0&0\\-3&3&3\\-3&3&3\end{bmatrix}\begin{bmatrix}3\\2\\1\end{bmatrix}
\displaystyle =\begin{bmatrix}0\\-9+6+3\\-9+6+3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }z=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }z=k\text{ in equations }(1)\text{ and }(2),\text{ we get}
\displaystyle x-y=3-k
\displaystyle 2x+y=2+k
\displaystyle \begin{bmatrix}1&-1\\2&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3-k\\2+k\end{bmatrix}
\displaystyle A_1X_1=B_1,\text{ where }A_1=\begin{bmatrix}1&-1\\2&1\end{bmatrix},\ X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B_1=\begin{bmatrix}3-k\\2+k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3\ne0
\displaystyle \text{adj }A_1=\begin{bmatrix}1&1\\-2&1\end{bmatrix}
\displaystyle A_1^{-1}=\frac{1}{|A_1|}\text{ adj }A_1=\frac{1}{3}\begin{bmatrix}1&1\\-2&1\end{bmatrix}
\displaystyle X_1=A_1^{-1}B_1
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{3}\begin{bmatrix}1&1\\-2&1\end{bmatrix}\begin{bmatrix}3-k\\2+k\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}3-k+2+k\\-6+2k+2+k\end{bmatrix}
\displaystyle =\begin{bmatrix}\dfrac{5}{3}\\[4pt]\dfrac{3k-4}{3}\end{bmatrix}
\displaystyle \therefore x=\frac{5}{3},\qquad y=\frac{3k-4}{3}\qquad\text{and}\qquad z=k,\quad k\in\mathbb{R}.
\displaystyle \text{These values of }x,\ y\text{ and }z\text{ also satisfy equation }(3).
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(v) }\quad x+y+z=6,\qquad x+2y+3z=14  , \qquad  x+4y+7z=30
\displaystyle \text{Answer:}
\displaystyle \text{(v)}
\displaystyle x+y+z=6\qquad\ldots(1)
\displaystyle x+2y+3z=14\qquad\ldots(2)
\displaystyle x+4y+7z=30\qquad\ldots(3)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\1&2&3\\1&4&7\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\14\\30\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\1&2&3\\1&4&7\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6\\14\\30\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&2&3\\1&4&7\end{vmatrix}
\displaystyle =1(14-12)-1(7-3)+1(4-2)
\displaystyle =2-4+2=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&3\\4&7\end{vmatrix}=2,\qquad C_{12}=-\begin{vmatrix}1&3\\1&7\end{vmatrix}=-4,\qquad C_{13}=\begin{vmatrix}1&2\\1&4\end{vmatrix}=2
\displaystyle C_{21}=-\begin{vmatrix}1&1\\4&7\end{vmatrix}=-3,\qquad C_{22}=\begin{vmatrix}1&1\\1&7\end{vmatrix}=6,\qquad C_{23}=-\begin{vmatrix}1&1\\1&4\end{vmatrix}=-3
\displaystyle C_{31}=\begin{vmatrix}1&1\\2&3\end{vmatrix}=1,\qquad C_{32}=-\begin{vmatrix}1&1\\1&3\end{vmatrix}=-2,\qquad C_{33}=\begin{vmatrix}1&1\\1&2\end{vmatrix}=1
\displaystyle \text{adj }A=\begin{bmatrix}2&-4&2\\-3&6&-3\\1&-2&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}2&-3&1\\-4&6&-2\\2&-3&1\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}2&-3&1\\-4&6&-2\\2&-3&1\end{bmatrix}\begin{bmatrix}6\\14\\30\end{bmatrix}
\displaystyle =\begin{bmatrix}12-42+30\\-24+84-60\\12-42+30\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }z=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }z=k\text{ in equations }(1)\text{ and }(2),\text{ we get}
\displaystyle x+y=6-k
\displaystyle x+2y=14-3k
\displaystyle \begin{bmatrix}1&1\\1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}6-k\\14-3k\end{bmatrix}
\displaystyle A_1X_1=B_1,\text{ where }A_1=\begin{bmatrix}1&1\\1&2\end{bmatrix},\ X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B_1=\begin{bmatrix}6-k\\14-3k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}1&1\\1&2\end{vmatrix}=2-1=1\ne0
\displaystyle \text{adj }A_1=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}
\displaystyle A_1^{-1}=\frac{1}{|A_1|}\text{ adj }A_1=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}
\displaystyle X_1=A_1^{-1}B_1
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}\begin{bmatrix}6-k\\14-3k\end{bmatrix}
\displaystyle =\begin{bmatrix}12-2k-14+3k\\-6+k+14-3k\end{bmatrix}
\displaystyle =\begin{bmatrix}k-2\\8-2k\end{bmatrix}
\displaystyle \therefore x=k-2,\qquad y=8-2k\qquad\text{and}\qquad z=k,\quad k\in\mathbb{R}.
\displaystyle \text{These values of }x,\ y\text{ and }z\text{ also satisfy equation }(3).
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that the following system of linear equations is consistent}
\displaystyle \text{and also find its solutions:}
\displaystyle \text{(vi) }\quad 2x+2y-2z=1,\qquad 4x+4y-z=2  , \qquad  6x+6y+2z=3
\displaystyle \text{Answer:}
\displaystyle \text{(vi)}
\displaystyle 2x+2y-2z=1\qquad\ldots(1)
\displaystyle 4x+4y-z=2\qquad\ldots(2)
\displaystyle 6x+6y+2z=3\qquad\ldots(3)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&2&-2\\4&4&-1\\6&6&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&2&-2\\4&4&-1\\6&6&2\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&2&-2\\4&4&-1\\6&6&2\end{vmatrix}
\displaystyle =2(8+6)-2(8+6)-2(24-24)
\displaystyle =28-28-0=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}4&-1\\6&2\end{vmatrix}=14,\qquad C_{12}=-\begin{vmatrix}4&-1\\6&2\end{vmatrix}=-14,\qquad C_{13}=\begin{vmatrix}4&4\\6&6\end{vmatrix}=0
\displaystyle C_{21}=-\begin{vmatrix}2&-2\\6&2\end{vmatrix}=-16,\qquad C_{22}=\begin{vmatrix}2&-2\\6&2\end{vmatrix}=16,\qquad C_{23}=-\begin{vmatrix}2&2\\6&6\end{vmatrix}=0
\displaystyle C_{31}=\begin{vmatrix}2&-2\\4&-1\end{vmatrix}=6,\qquad C_{32}=-\begin{vmatrix}2&-2\\4&-1\end{vmatrix}=-6,\qquad C_{33}=\begin{vmatrix}2&2\\4&4\end{vmatrix}=0
\displaystyle \text{adj }A=\begin{bmatrix}14&-14&0\\-16&16&0\\6&-6&0\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}14&-16&6\\-14&16&-6\\0&0&0\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}14&-16&6\\-14&16&-6\\0&0&0\end{bmatrix}\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle =\begin{bmatrix}14-32+18\\-14+32-18\\0\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B=0,
\displaystyle \text{the given system is consistent and has infinitely many solutions.}
\displaystyle \text{Let }y=k,\text{ where }k\in\mathbb{R}.
\displaystyle \text{Substituting }y=k\text{ in equations }(1)\text{ and }(2),\text{ we get}
\displaystyle 2x-2z=1-2k
\displaystyle 4x-z=2-4k
\displaystyle \begin{bmatrix}2&-2\\4&-1\end{bmatrix}\begin{bmatrix}x\\z\end{bmatrix}=\begin{bmatrix}1-2k\\2-4k\end{bmatrix}
\displaystyle A_1X_1=B_1,\text{ where }A_1=\begin{bmatrix}2&-2\\4&-1\end{bmatrix},\ X_1=\begin{bmatrix}x\\z\end{bmatrix}\text{ and }B_1=\begin{bmatrix}1-2k\\2-4k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}2&-2\\4&-1\end{vmatrix}=-2+8=6\ne0
\displaystyle \text{adj }A_1=\begin{bmatrix}-1&2\\-4&2\end{bmatrix}
\displaystyle A_1^{-1}=\frac{1}{|A_1|}\text{ adj }A_1=\frac{1}{6}\begin{bmatrix}-1&2\\-4&2\end{bmatrix}
\displaystyle X_1=A_1^{-1}B_1
\displaystyle \begin{bmatrix}x\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}-1&2\\-4&2\end{bmatrix}\begin{bmatrix}1-2k\\2-4k\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}-1+2k+4-8k\\-4+8k+4-8k\end{bmatrix}
\displaystyle =\begin{bmatrix}\dfrac{3-6k}{6}\\0\end{bmatrix}=\begin{bmatrix}\dfrac{1-2k}{2}\\0\end{bmatrix}
\displaystyle \therefore x=\frac{1-2k}{2},\qquad y=k\qquad\text{and}\qquad z=0,\quad k\in\mathbb{R}.
\displaystyle \text{These values of }x,\ y\text{ and }z\text{ also satisfy equation }(3).
\displaystyle \text{Hence, the given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(i) }\quad 2x+5y=7,\qquad 6x+15y=13
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&5\\6&15\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\13\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&5\\6&15\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}7\\13\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&5\\6&15\end{vmatrix}=30-30=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=15,\qquad C_{12}=-6,\qquad C_{21}=-5,\qquad C_{22}=2
\displaystyle \text{adj }A=\begin{bmatrix}15&-6\\-5&2\end{bmatrix}^{T}=\begin{bmatrix}15&-5\\-6&2\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}15&-5\\-6&2\end{bmatrix}\begin{bmatrix}7\\13\end{bmatrix}
\displaystyle =\begin{bmatrix}105-65\\-42+26\end{bmatrix}=\begin{bmatrix}40\\-16\end{bmatrix}\ne\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(ii) }\quad 2x+3y=5,\qquad 6x+9y=10
\displaystyle \text{Answer:}
\displaystyle \text{(ii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&3\\6&9\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\10\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&3\\6&9\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}5\\10\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3\\6&9\end{vmatrix}=18-18=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=9,\qquad C_{12}=-6,\qquad C_{21}=-3,\qquad C_{22}=2
\displaystyle \text{adj }A=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^{T}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\10\end{bmatrix}
\displaystyle =\begin{bmatrix}45-30\\-30+20\end{bmatrix}=\begin{bmatrix}15\\-10\end{bmatrix}\ne\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(iii) }\quad 4x-2y=3,\qquad 6x-3y=5
\displaystyle \text{Answer:}
\displaystyle \text{(iii)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}4&-2\\6&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}4&-2\\6&-3\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}4&-2\\6&-3\end{vmatrix}=-12+12=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=-3,\qquad C_{12}=-6,\qquad C_{21}=2,\qquad C_{22}=4
\displaystyle \text{adj }A=\begin{bmatrix}-3&-6\\2&4\end{bmatrix}^{T}=\begin{bmatrix}-3&2\\-6&4\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}-3&2\\-6&4\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}
\displaystyle =\begin{bmatrix}-9+10\\-18+20\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(iv) }\quad 4x-5y-2z=2,\qquad 5x-4y+2z=-2
\displaystyle 2x+2y+8z=-1
\displaystyle \text{Answer:}
\displaystyle \text{(iv)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}4&-5&-2\\5&-4&2\\2&2&8\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-2\\-1\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}4&-5&-2\\5&-4&2\\2&2&8\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\-2\\-1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}4&-5&-2\\5&-4&2\\2&2&8\end{vmatrix}
\displaystyle =4(-32-4)+5(40-4)-2(10+8)
\displaystyle =-144+180-36=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-4&2\\2&8\end{vmatrix}=-36,\qquad C_{12}=-\begin{vmatrix}5&2\\2&8\end{vmatrix}=-36,\qquad C_{13}=\begin{vmatrix}5&-4\\2&2\end{vmatrix}=18
\displaystyle C_{21}=-\begin{vmatrix}-5&-2\\2&8\end{vmatrix}=36,\qquad C_{22}=\begin{vmatrix}4&-2\\2&8\end{vmatrix}=36,\qquad C_{23}=-\begin{vmatrix}4&-5\\2&2\end{vmatrix}=-18
\displaystyle C_{31}=\begin{vmatrix}-5&-2\\-4&2\end{vmatrix}=-18,\qquad C_{32}=-\begin{vmatrix}4&-2\\5&2\end{vmatrix}=-18,\qquad C_{33}=\begin{vmatrix}4&-5\\5&-4\end{vmatrix}=9
\displaystyle \text{adj }A=\begin{bmatrix}-36&-36&18\\36&36&-18\\-18&-18&9\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-36&36&-18\\-36&36&-18\\18&-18&9\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}-36&36&-18\\-36&36&-18\\18&-18&9\end{bmatrix}\begin{bmatrix}2\\-2\\-1\end{bmatrix}
\displaystyle =\begin{bmatrix}-72-72+18\\-72-72+18\\36+36-9\end{bmatrix}
\displaystyle =\begin{bmatrix}-126\\-126\\63\end{bmatrix}\ne\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(v) }\quad 3x-y-2z=2,\qquad 2y-z=-1,\qquad 3x-5y=3
\displaystyle \text{Answer:}
\displaystyle \text{(v)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\-1\\3\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}
\displaystyle =3(0-5)+1(0+3)-2(0-6)
\displaystyle =-15+3+12=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&-1\\-5&0\end{vmatrix}=-5,\qquad C_{12}=-\begin{vmatrix}0&-1\\3&0\end{vmatrix}=-3,\qquad C_{13}=\begin{vmatrix}0&2\\3&-5\end{vmatrix}=-6
\displaystyle C_{21}=-\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}=10,\qquad C_{22}=\begin{vmatrix}3&-2\\3&0\end{vmatrix}=6,\qquad C_{23}=-\begin{vmatrix}3&-1\\3&-5\end{vmatrix}=12
\displaystyle C_{31}=\begin{vmatrix}-1&-2\\2&-1\end{vmatrix}=5,\qquad C_{32}=-\begin{vmatrix}3&-2\\0&-1\end{vmatrix}=3,\qquad C_{33}=\begin{vmatrix}3&-1\\0&2\end{vmatrix}=6
\displaystyle \text{adj }A=\begin{bmatrix}-5&-3&-6\\10&6&12\\5&3&6\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}\begin{bmatrix}2\\-1\\3\end{bmatrix}
\displaystyle =\begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix}
\displaystyle =\begin{bmatrix}-5\\-3\\-6\end{bmatrix}\ne\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the following system of linear equations is inconsistent:}
\displaystyle \text{(vi) }\quad x+y-2z=5,\qquad x-2y+z=-2,\qquad -2x+y+z=4
\displaystyle \text{Answer:}
\displaystyle \text{(vi)}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&-2\\1&-2&1\\-2&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-2\\4\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&-2\\1&-2&1\\-2&1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}5\\-2\\4\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&-2\\1&-2&1\\-2&1&1\end{vmatrix}
\displaystyle =1(-2-1)-1(1+2)-2(1-4)
\displaystyle =-3-3+6=0
\displaystyle \therefore A\text{ is a singular matrix.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-2&1\\1&1\end{vmatrix}=-3,\qquad C_{12}=-\begin{vmatrix}1&1\\-2&1\end{vmatrix}=-3,\qquad C_{13}=\begin{vmatrix}1&-2\\-2&1\end{vmatrix}=-3
\displaystyle C_{21}=-\begin{vmatrix}1&-2\\1&1\end{vmatrix}=-3,\qquad C_{22}=\begin{vmatrix}1&-2\\-2&1\end{vmatrix}=-3,\qquad C_{23}=-\begin{vmatrix}1&1\\-2&1\end{vmatrix}=-3
\displaystyle C_{31}=\begin{vmatrix}1&-2\\-2&1\end{vmatrix}=-3,\qquad C_{32}=-\begin{vmatrix}1&-2\\1&1\end{vmatrix}=-3,\qquad C_{33}=\begin{vmatrix}1&1\\1&-2\end{vmatrix}=-3
\displaystyle \text{adj }A=\begin{bmatrix}-3&-3&-3\\-3&-3&-3\\-3&-3&-3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-3&-3&-3\\-3&-3&-3\\-3&-3&-3\end{bmatrix}
\displaystyle (\text{adj }A)B=\begin{bmatrix}-3&-3&-3\\-3&-3&-3\\-3&-3&-3\end{bmatrix}\begin{bmatrix}5\\-2\\4\end{bmatrix}
\displaystyle =\begin{bmatrix}-15+6-12\\-15+6-12\\-15+6-12\end{bmatrix}
\displaystyle =\begin{bmatrix}-21\\-21\\-21\end{bmatrix}\ne\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{Since }|A|=0\text{ and }(\text{adj }A)B\ne0,
\displaystyle \text{the given system of equations is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\text{ and }B=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle \text{are two square matrices, find }AB\text{ and hence solve the system of linear equations:}
\displaystyle x-y=3,\qquad 2x+3y+4z=17,\qquad y+2z=7.\qquad \text{[CBSE 2010, 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }A=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\text{ and }B=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle =\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}
\displaystyle =\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I_3
\displaystyle \therefore A\left(\frac{1}{6}B\right)=I_3
\displaystyle \text{Since }A\text{ and }B\text{ are square matrices, }A^{-1}=\frac{1}{6}B
\displaystyle \therefore A^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle AX=C,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }C=\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle \therefore X=A^{-1}C
\displaystyle X=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}
\displaystyle \therefore x=2,\qquad y=-1,\qquad z=4
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\text{ find }A^{-1}\text{ and hence solve the system}
\displaystyle \text{of linear equations }2x-3y+5z=11,\quad 3x+2y-4z=-5,\quad x+y-2z=-3.
\displaystyle \hspace{2cm}\text{[CBSE 2007, 2009, 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }A=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix}
\displaystyle =2(-4+4)+3(-6+4)+5(3-2)
\displaystyle =0-6+5=-1\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&-4\\1&-2\end{vmatrix}=0,\qquad C_{12}=-\begin{vmatrix}3&-4\\1&-2\end{vmatrix}=2,\qquad C_{13}=\begin{vmatrix}3&2\\1&1\end{vmatrix}=1
\displaystyle C_{21}=-\begin{vmatrix}-3&5\\1&-2\end{vmatrix}=-1,\qquad C_{22}=\begin{vmatrix}2&5\\1&-2\end{vmatrix}=-9,\qquad C_{23}=-\begin{vmatrix}2&-3\\1&1\end{vmatrix}=-5
\displaystyle C_{31}=\begin{vmatrix}-3&5\\2&-4\end{vmatrix}=2,\qquad C_{32}=-\begin{vmatrix}2&5\\3&-4\end{vmatrix}=23,\qquad C_{33}=\begin{vmatrix}2&-3\\3&2\end{vmatrix}=13
\displaystyle \text{adj }A=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}
\displaystyle =\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}
\displaystyle AX=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}
\displaystyle =\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}
\displaystyle =\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \therefore x=1,\qquad y=2,\qquad z=3
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find }A^{-1},\text{ if }A=\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}.\text{ Hence, solve the following}
\displaystyle \text{system of linear equations: }x+2y+5z=10,\quad x-y-z=-2,\quad 2x+3y-z=-11.
\displaystyle \hspace{4cm}\text{[CBSE 2010, 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }A=\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{vmatrix}
\displaystyle =1(1+3)-2(-1+2)+5(3+2)
\displaystyle =4-2+25=27\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-1&-1\\3&-1\end{vmatrix}=4,\qquad C_{12}=-\begin{vmatrix}1&-1\\2&-1\end{vmatrix}=-1,\qquad C_{13}=\begin{vmatrix}1&-1\\2&3\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}2&5\\3&-1\end{vmatrix}=17,\qquad C_{22}=\begin{vmatrix}1&5\\2&-1\end{vmatrix}=-11,\qquad C_{23}=-\begin{vmatrix}1&2\\2&3\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}2&5\\-1&-1\end{vmatrix}=3,\qquad C_{32}=-\begin{vmatrix}1&5\\1&-1\end{vmatrix}=6,\qquad C_{33}=\begin{vmatrix}1&2\\1&-1\end{vmatrix}=-3
\displaystyle \text{adj }A=\begin{bmatrix}4&-1&5\\17&-11&1\\3&6&-3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{27}\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\-2\\-11\end{bmatrix}
\displaystyle AX=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}10\\-2\\-11\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{27}\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}\begin{bmatrix}10\\-2\\-11\end{bmatrix}
\displaystyle =\frac{1}{27}\begin{bmatrix}40-34-33\\-10+22-66\\50-2+33\end{bmatrix}
\displaystyle =\frac{1}{27}\begin{bmatrix}-27\\-54\\81\end{bmatrix}=\begin{bmatrix}-1\\-2\\3\end{bmatrix}
\displaystyle \therefore x=-1,\qquad y=-2,\qquad z=3
\displaystyle \\

 

\displaystyle \textbf{Question 8:}
\displaystyle \text{(i) If }A=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix},\text{ find }A^{-1}.\text{ Using }A^{-1},\text{ solve the system of linear equations}
\displaystyle x-2y=10,\qquad 2x+y+3z=8,\qquad -2y+z=7.
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Here, }A=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&-2&0\\2&1&3\\0&-2&1\end{vmatrix}
\displaystyle =1(1+6)+2(2-0)+0(-4-0)
\displaystyle =7+4=11\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}1&3\\-2&1\end{vmatrix}=7,\qquad C_{12}=-\begin{vmatrix}2&3\\0&1\end{vmatrix}=-2,\qquad C_{13}=\begin{vmatrix}2&1\\0&-2\end{vmatrix}=-4
\displaystyle C_{21}=-\begin{vmatrix}-2&0\\-2&1\end{vmatrix}=2,\qquad C_{22}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,\qquad C_{23}=-\begin{vmatrix}1&-2\\0&-2\end{vmatrix}=2
\displaystyle C_{31}=\begin{vmatrix}-2&0\\1&3\end{vmatrix}=-6,\qquad C_{32}=-\begin{vmatrix}1&0\\2&3\end{vmatrix}=-3,\qquad C_{33}=\begin{vmatrix}1&-2\\2&1\end{vmatrix}=5
\displaystyle \therefore \text{adj }A=\begin{bmatrix}7&-2&-4\\2&1&2\\-6&-3&5\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle AX=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}70+16-42\\-20+8-21\\-40+16+35\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}44\\-33\\11\end{bmatrix}=\begin{bmatrix}4\\-3\\1\end{bmatrix}
\displaystyle \therefore x=4,\qquad y=-3,\qquad z=1
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(ii) If }A=\begin{bmatrix}3&-4&2\\2&3&5\\1&0&1\end{bmatrix},\text{ find }A^{-1}\text{ and hence solve the following system of linear equations:}
\displaystyle 3x-4y+2z=-1,\qquad 2x+3y+5z=7,\qquad x+z=2.\qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{(ii)}
\displaystyle \text{Here, }A=\begin{bmatrix}3&-4&2\\2&3&5\\1&0&1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&-4&2\\2&3&5\\1&0&1\end{vmatrix}
\displaystyle =3(3-0)+4(2-5)+2(0-3)
\displaystyle =9-12-6=-9\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}3&5\\0&1\end{vmatrix}=3,\qquad C_{12}=-\begin{vmatrix}2&5\\1&1\end{vmatrix}=3,\qquad C_{13}=\begin{vmatrix}2&3\\1&0\end{vmatrix}=-3
\displaystyle C_{21}=-\begin{vmatrix}-4&2\\0&1\end{vmatrix}=4,\qquad C_{22}=\begin{vmatrix}3&2\\1&1\end{vmatrix}=1,\qquad C_{23}=-\begin{vmatrix}3&-4\\1&0\end{vmatrix}=-4
\displaystyle C_{31}=\begin{vmatrix}-4&2\\3&5\end{vmatrix}=-26,\qquad C_{32}=-\begin{vmatrix}3&2\\2&5\end{vmatrix}=-11,\qquad C_{33}=\begin{vmatrix}3&-4\\2&3\end{vmatrix}=17
\displaystyle \text{adj }A=\begin{bmatrix}3&3&-3\\4&1&-4\\-26&-11&17\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}3&4&-26\\3&1&-11\\-3&-4&17\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{9}\begin{bmatrix}3&4&-26\\3&1&-11\\-3&-4&17\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&-4&2\\2&3&5\\1&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-1\\7\\2\end{bmatrix}
\displaystyle AX=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}-1\\7\\2\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{9}\begin{bmatrix}3&4&-26\\3&1&-11\\-3&-4&17\end{bmatrix}\begin{bmatrix}-1\\7\\2\end{bmatrix}
\displaystyle =-\frac{1}{9}\begin{bmatrix}-3+28-52\\-3+7-22\\3-28+34\end{bmatrix}
\displaystyle =-\frac{1}{9}\begin{bmatrix}-27\\-18\\9\end{bmatrix}=\begin{bmatrix}3\\2\\-1\end{bmatrix}
\displaystyle \therefore x=3,\qquad y=2,\qquad z=-1
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(iii) }A=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix},\text{ find }AB.\text{ Hence, solve the}
\displaystyle \text{system of equations: }x-2y=10,\quad 2x+y+3z=8,\quad -2y+z=7.\qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{(iii)}
\displaystyle \text{Here, }A=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}7+4+0&2-2+0&-6+6+0\\14-2-12&4+1+6&-12-3+15\\0+4-4&0-2+2&0+6+5\end{bmatrix}
\displaystyle =\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}=11I_3
\displaystyle \therefore A\left(\frac{1}{11}B\right)=I_3
\displaystyle \text{Since }A\text{ and }B\text{ are square matrices, }A^{-1}=\frac{1}{11}B
\displaystyle \therefore A^{-1}=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle AX=C,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }C=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle \therefore X=A^{-1}C
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}70+16-42\\-20+8-21\\-40+16+35\end{bmatrix}
\displaystyle =\frac{1}{11}\begin{bmatrix}44\\-33\\11\end{bmatrix}=\begin{bmatrix}4\\-3\\1\end{bmatrix}
\displaystyle \therefore x=4,\qquad y=-3,\qquad z=1
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(iv) If }A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix},\text{ find }A^{-1}.\text{ Using }A^{-1},\text{ solve the system of linear}
\displaystyle \text{equations }x-2y=10,\qquad 2x-y-z=8,\qquad -2y+z=7.
\displaystyle \text{Answer:}
\displaystyle \text{(iv)}
\displaystyle \text{Here, }A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{vmatrix}
\displaystyle =1(-1-2)-2(-2-0)+0
\displaystyle =-3+4=1\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-1&-2\\-1&1\end{vmatrix}=-3,\qquad C_{12}=-\begin{vmatrix}-2&-2\\0&1\end{vmatrix}=2,\qquad C_{13}=\begin{vmatrix}-2&-1\\0&-1\end{vmatrix}=2
\displaystyle C_{21}=-\begin{vmatrix}2&0\\-1&1\end{vmatrix}=-2,\qquad C_{22}=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1,\qquad C_{23}=-\begin{vmatrix}1&2\\0&-1\end{vmatrix}=1
\displaystyle C_{31}=\begin{vmatrix}2&0\\-1&-2\end{vmatrix}=-4,\qquad C_{32}=-\begin{vmatrix}1&0\\-2&-2\end{vmatrix}=2,\qquad C_{33}=\begin{vmatrix}1&2\\-2&-1\end{vmatrix}=3
\displaystyle \therefore \text{adj }A=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}
\displaystyle \text{The coefficient matrix of the given system is}
\displaystyle C=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}=A^{T}
\displaystyle \text{We know that }(A^{T})^{-1}=(A^{-1})^{T}.
\displaystyle \therefore C^{-1}=(A^{-1})^{T}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle CX=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle \therefore X=C^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}
\displaystyle =\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}
\displaystyle =\begin{bmatrix}0\\-5\\-3\end{bmatrix}
\displaystyle \therefore x=0,\qquad y=-5,\qquad z=-3
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(v) Given }A=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\quad B=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\text{ find }BA\text{ and use this to}
\displaystyle \text{solve the system of equations }y+2z=7,\qquad x-y=3,\qquad 2x+3y+4z=17.
\displaystyle \text{Answer:}
\displaystyle \text{(v)}
\displaystyle \text{Here, }A=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\text{ and }B=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}
\displaystyle BA=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle =\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}
\displaystyle =\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I_3
\displaystyle \therefore B\left(\frac{1}{6}A\right)=I_3
\displaystyle \text{Since }A\text{ and }B\text{ are square matrices, }B^{-1}=\frac{1}{6}A
\displaystyle \therefore B^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}
\displaystyle \text{Rearranging the given equations in the order corresponding to the rows of }B,
\displaystyle x-y=3,\qquad 2x+3y+4z=17,\qquad y+2z=7
\displaystyle \text{The system can be written in matrix form as}
\displaystyle \begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle BX=C,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }C=\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle \therefore X=B^{-1}C
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}
\displaystyle \therefore x=2,\qquad y=-1,\qquad z=4
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(vi) If }A=\begin{bmatrix}2&3&1\\1&2&2\\-3&1&-1\end{bmatrix},\text{ find }A^{-1}\text{ and hence solve the system of equations}
\displaystyle 2x+y-3z=13,\qquad 3x+2y+z=4,\qquad x+2y-z=8.\qquad \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{(vi)}
\displaystyle \text{Here, }A=\begin{bmatrix}2&3&1\\1&2&2\\-3&1&-1\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3&1\\1&2&2\\-3&1&-1\end{vmatrix}
\displaystyle =2(-2-2)-3(-1+6)+(1+6)
\displaystyle =-8-15+7=-16\ne0
\displaystyle \therefore A\text{ is invertible.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&2\\1&-1\end{vmatrix}=-4,\qquad C_{12}=-\begin{vmatrix}1&2\\-3&-1\end{vmatrix}=-5,\qquad C_{13}=\begin{vmatrix}1&2\\-3&1\end{vmatrix}=7
\displaystyle C_{21}=-\begin{vmatrix}3&1\\1&-1\end{vmatrix}=4,\qquad C_{22}=\begin{vmatrix}2&1\\-3&-1\end{vmatrix}=1,\qquad C_{23}=-\begin{vmatrix}2&3\\-3&1\end{vmatrix}=-11
\displaystyle C_{31}=\begin{vmatrix}3&1\\2&2\end{vmatrix}=4,\qquad C_{32}=-\begin{vmatrix}2&1\\1&2\end{vmatrix}=-3,\qquad C_{33}=\begin{vmatrix}2&3\\1&2\end{vmatrix}=1
\displaystyle \therefore \text{adj }A=\begin{bmatrix}-4&-5&7\\4&1&-11\\4&-3&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{16}\begin{bmatrix}-4&4&4\\-5&1&-3\\7&-11&1\end{bmatrix}
\displaystyle \text{The coefficient matrix of the given system is }A^{T}.
\displaystyle \therefore A^{T}X=B,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}13\\4\\8\end{bmatrix}
\displaystyle |A^{T}|=|A|=-16\ne0
\displaystyle \therefore \text{the system has a unique solution.}
\displaystyle X=(A^{T})^{-1}B=(A^{-1})^{T}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{16}\begin{bmatrix}-4&-5&7\\4&1&-11\\4&-3&1\end{bmatrix}\begin{bmatrix}13\\4\\8\end{bmatrix}
\displaystyle =-\frac{1}{16}\begin{bmatrix}-52-20+56\\52+4-88\\52-12+8\end{bmatrix}
\displaystyle =-\frac{1}{16}\begin{bmatrix}-16\\-32\\48\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}
\displaystyle \therefore x=1,\qquad y=2,\qquad z=-3
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(vii) Use the product of }\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\text{ to solve the system of equations}
\displaystyle x+3z=9,\qquad -x+2y-2z=4,\qquad 2x-3y+4z=-3.\qquad \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{(vii)}
\displaystyle \text{Let }A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2-9+12&0-2+2&1+3-4\\0+18-18&0+4-3&0-6+6\\-6-18+24&0-4+4&3+6-8\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3
\displaystyle \therefore B=A^{-1}\qquad (1)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}
\displaystyle \text{Since }\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}=A^{T},
\displaystyle A^{T}X=C,\text{ where }X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }C=\begin{bmatrix}9\\4\\-3\end{bmatrix}
\displaystyle \therefore X=(A^{T})^{-1}C=(A^{-1})^{T}C=B^{T}C\qquad\text{[Using (1)]}
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}^{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&9&6\\0&2&1\\1&-3&-2\end{bmatrix}\begin{bmatrix}9\\4\\-3\end{bmatrix}
\displaystyle =\begin{bmatrix}-18+36-18\\0+8-3\\9-12+6\end{bmatrix}=\begin{bmatrix}0\\5\\3\end{bmatrix}
\displaystyle \therefore x=0,\qquad y=5,\qquad z=3
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The sum of three numbers is }2.\text{ If twice the second number is added to the sum of the}
\displaystyle \text{first and third, the sum is }1.\text{ By adding the second and third numbers to five times the first}
\displaystyle \text{number, we get }6.\text{ Find the three numbers by using matrices.}\qquad \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three numbers be }x,\ y\text{ and }z.
\displaystyle \text{According to the question,}
\displaystyle x+y+z=2
\displaystyle x+2y+z=1
\displaystyle 5x+y+z=6
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\6\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}2\\1\\6\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&2&1\\5&1&1\end{vmatrix}
\displaystyle =1(2-1)-1(1-5)+1(1-10)
\displaystyle =1+4-9=-4\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}2&1\\1&1\end{vmatrix}=1,\qquad C_{12}=-\begin{vmatrix}1&1\\5&1\end{vmatrix}=4,\qquad C_{13}=\begin{vmatrix}1&2\\5&1\end{vmatrix}=-9
\displaystyle C_{21}=-\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,\qquad C_{22}=\begin{vmatrix}1&1\\5&1\end{vmatrix}=-4,\qquad C_{23}=-\begin{vmatrix}1&1\\5&1\end{vmatrix}=4
\displaystyle C_{31}=\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,\qquad C_{32}=-\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,\qquad C_{33}=\begin{vmatrix}1&1\\1&2\end{vmatrix}=1
\displaystyle \text{adj }A=\begin{bmatrix}1&4&-9\\0&-4&4\\-1&0&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{4}\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{4}\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}\begin{bmatrix}2\\1\\6\end{bmatrix}
\displaystyle =-\frac{1}{4}\begin{bmatrix}2+0-6\\8-4+0\\-18+4+6\end{bmatrix}
\displaystyle =-\frac{1}{4}\begin{bmatrix}-4\\4\\-8\end{bmatrix}=\begin{bmatrix}1\\-1\\2\end{bmatrix}
\displaystyle \therefore x=1,\qquad y=-1,\qquad z=2
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{An amount of Rs }10{,}000\text{ is put into three investments at the rates of }10\%,\ 12\%
\displaystyle \text{and }15\%\text{ per annum. The combined income is Rs }1310,\text{ and the combined income of the first}
\displaystyle \text{and second investments is Rs }190\text{ short of the income from the third investment. Find the}
\displaystyle \text{investment in each using the matrix method.}\qquad \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x,\ y\text{ and }z\text{ be the investments at the rates of }10\%,\ 12\%\text{ and }15\%\text{ per annum, respectively.}
\displaystyle \text{Total investment}=\text{Rs }10{,}000
\displaystyle \therefore x+y+z=10{,}000
\displaystyle \text{Income from the first investment of Rs }x=\text{Rs }\frac{10x}{100}=\text{Rs }0.1x
\displaystyle \text{Income from the second investment of Rs }y=\text{Rs }\frac{12y}{100}=\text{Rs }0.12y
\displaystyle \text{Income from the third investment of Rs }z=\text{Rs }\frac{15z}{100}=\text{Rs }0.15z
\displaystyle \therefore \text{total annual income}=\text{Rs }(0.1x+0.12y+0.15z)
\displaystyle 0.1x+0.12y+0.15z=1310
\displaystyle \text{Also, the combined income from the first two investments is Rs }190\text{ short of the income from the third investment.}
\displaystyle \therefore 0.1x+0.12y=0.15z-190
\displaystyle \Rightarrow -0.1x-0.12y+0.15z=190
\displaystyle \text{Thus, the system of simultaneous linear equations is}
\displaystyle x+y+z=10{,}000
\displaystyle 0.1x+0.12y+0.15z=1310
\displaystyle -0.1x-0.12y+0.15z=190
\displaystyle \text{The system can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\0.1&0.12&0.15\\-0.1&-0.12&0.15\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10{,}000\\1310\\190\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\0.1&0.12&0.15\\-0.1&-0.12&0.15\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}10{,}000\\1310\\190\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\0.1&0.12&0.15\\-0.1&-0.12&0.15\end{vmatrix}
\displaystyle =1(0.15\times0.12+0.15\times0.12)-1(0.15\times0.1+0.15\times0.1)
\displaystyle \qquad +1(-0.1\times0.12+0.12\times0.1)
\displaystyle =0.036-0.03+0=0.006\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}0.12&0.15\\-0.12&0.15\end{vmatrix}=0.036,\qquad C_{12}=-\begin{vmatrix}0.1&0.15\\-0.1&0.15\end{vmatrix}=-0.03,\qquad C_{13}=\begin{vmatrix}0.1&0.12\\-0.1&-0.12\end{vmatrix}=0
\displaystyle C_{21}=-\begin{vmatrix}1&1\\-0.12&0.15\end{vmatrix}=-0.27,\qquad C_{22}=\begin{vmatrix}1&1\\-0.1&0.15\end{vmatrix}=0.25,\qquad C_{23}=-\begin{vmatrix}1&1\\-0.1&-0.12\end{vmatrix}=0.02
\displaystyle C_{31}=\begin{vmatrix}1&1\\0.12&0.15\end{vmatrix}=0.03,\qquad C_{32}=-\begin{vmatrix}1&1\\0.1&0.15\end{vmatrix}=-0.05,\qquad C_{33}=\begin{vmatrix}1&1\\0.1&0.12\end{vmatrix}=0.02
\displaystyle \text{adj }A=\begin{bmatrix}0.036&-0.03&0\\-0.27&0.25&0.02\\0.03&-0.05&0.02\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}0.036&-0.27&0.03\\-0.03&0.25&-0.05\\0&0.02&0.02\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{0.006}\begin{bmatrix}0.036&-0.27&0.03\\-0.03&0.25&-0.05\\0&0.02&0.02\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{0.006}\begin{bmatrix}0.036&-0.27&0.03\\-0.03&0.25&-0.05\\0&0.02&0.02\end{bmatrix}\begin{bmatrix}10{,}000\\1310\\190\end{bmatrix}
\displaystyle =\frac{1}{0.006}\begin{bmatrix}360-353.7+5.7\\-300+327.5-9.5\\0+26.2+3.8\end{bmatrix}
\displaystyle =\frac{1}{0.006}\begin{bmatrix}12\\18\\30\end{bmatrix}=\begin{bmatrix}2000\\3000\\5000\end{bmatrix}
\displaystyle \therefore x=2000,\qquad y=3000,\qquad z=5000
\displaystyle \text{Thus, the three investments are Rs }2000,\text{ Rs }3000\text{ and Rs }5000,\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A company produces three products every day. Their total production on a certain day is}
\displaystyle \text{ }45\text{ tons. It is found that the production of the third product exceeds the production of the}
\displaystyle \text{first product by }8\text{ tons, while the total production of the first and third products is}
\displaystyle \text{twice the production of the second product. Determine the production level of each}
\displaystyle \text{product using the matrix method.}\qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x,\ y\text{ and }z\text{ be the production levels of the first, second and third products, respectively.}
\displaystyle \text{According to the question,}
\displaystyle x+y+z=45\qquad (1)
\displaystyle -x+z=8\qquad (2)
\displaystyle x+z=2y
\displaystyle \Rightarrow x-2y+z=0\qquad (3)
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\-1&0&1\\1&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}45\\8\\0\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\-1&0&1\\1&-2&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}45\\8\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\-1&0&1\\1&-2&1\end{vmatrix}
\displaystyle =1(0+2)-1(-1-1)+1(2-0)
\displaystyle =2+2+2=6\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}0&1\\-2&1\end{vmatrix}=2,\qquad C_{12}=-\begin{vmatrix}-1&1\\1&1\end{vmatrix}=2,\qquad C_{13}=\begin{vmatrix}-1&0\\1&-2\end{vmatrix}=2
\displaystyle C_{21}=-\begin{vmatrix}1&1\\-2&1\end{vmatrix}=-3,\qquad C_{22}=\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,\qquad C_{23}=-\begin{vmatrix}1&1\\1&-2\end{vmatrix}=3
\displaystyle C_{31}=\begin{vmatrix}1&1\\0&1\end{vmatrix}=1,\qquad C_{32}=-\begin{vmatrix}1&1\\-1&1\end{vmatrix}=-2,\qquad C_{33}=\begin{vmatrix}1&1\\-1&0\end{vmatrix}=1
\displaystyle \text{adj }A=\begin{bmatrix}2&2&2\\-3&0&3\\1&-2&1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{6}\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}\begin{bmatrix}45\\8\\0\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}90-24+0\\90+0+0\\90+24+0\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}66\\90\\114\end{bmatrix}=\begin{bmatrix}11\\15\\19\end{bmatrix}
\displaystyle \therefore x=11,\qquad y=15,\qquad z=19
\displaystyle \text{Thus, the production levels of the first, second and third products are }11,\ 15\text{ and }19\text{ tons, respectively.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The prices of three commodities }P,\ Q\text{ and }R\text{ are Rs }x,\ y\text{ and }z\text{ per unit, respectively.}
\displaystyle \text{A purchases }4\text{ units of }R\text{ and sells }3\text{ units of }P\text{ and }5\text{ units of }Q.\text{ B purchases }3\text{ units}
\displaystyle \text{of }Q\text{ and sells }2\text{ units of }P\text{ and }1\text{ unit of }R.\text{ C purchases }1\text{ unit of }P\text{ and sells }4\text{ units}
\displaystyle \text{of }Q\text{ and }6\text{ units of }R.\text{ In the process, A, B and C earn Rs }6000,\text{ Rs }5000\text{ and Rs }13000,\text{ respectively.}
\displaystyle \text{If selling the units is positive earning and buying the units is negative earning, find the}
\displaystyle \text{price per unit of the three commodities using the matrix method.}\qquad \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The prices of the commodities }P,\ Q\text{ and }R\text{ are Rs }x,\text{ Rs }y\text{ and Rs }z\text{ per unit, respectively.}
\displaystyle \text{According to the question,}
\displaystyle 3x+5y-4z=6000
\displaystyle 2x-3y+z=5000
\displaystyle -x+4y+6z=13000
\displaystyle \text{The given system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&5&-4\\2&-3&1\\-1&4&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6000\\5000\\13000\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}3&5&-4\\2&-3&1\\-1&4&6\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6000\\5000\\13000\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&5&-4\\2&-3&1\\-1&4&6\end{vmatrix}
\displaystyle =3(-18-4)-5(12+1)-4(8-3)
\displaystyle =-66-65-20=-151\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=\begin{vmatrix}-3&1\\4&6\end{vmatrix}=-22,\qquad C_{12}=-\begin{vmatrix}2&1\\-1&6\end{vmatrix}=-13,\qquad C_{13}=\begin{vmatrix}2&-3\\-1&4\end{vmatrix}=5
\displaystyle C_{21}=-\begin{vmatrix}5&-4\\4&6\end{vmatrix}=-46,\qquad C_{22}=\begin{vmatrix}3&-4\\-1&6\end{vmatrix}=14,\qquad C_{23}=-\begin{vmatrix}3&5\\-1&4\end{vmatrix}=-17
\displaystyle C_{31}=\begin{vmatrix}5&-4\\-3&1\end{vmatrix}=-7,\qquad C_{32}=-\begin{vmatrix}3&-4\\2&1\end{vmatrix}=-11,\qquad C_{33}=\begin{vmatrix}3&5\\2&-3\end{vmatrix}=-19
\displaystyle \text{adj }A=\begin{bmatrix}-22&-13&5\\-46&14&-17\\-7&-11&-19\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{151}\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{151}\begin{bmatrix}-22&-46&-7\\-13&14&-11\\5&-17&-19\end{bmatrix}\begin{bmatrix}6000\\5000\\13000\end{bmatrix}
\displaystyle =-\frac{1}{151}\begin{bmatrix}-132000-230000-91000\\-78000+70000-143000\\30000-85000-247000\end{bmatrix}
\displaystyle =-\frac{1}{151}\begin{bmatrix}-453000\\-151000\\-302000\end{bmatrix}=\begin{bmatrix}3000\\1000\\2000\end{bmatrix}
\displaystyle \therefore x=3000,\qquad y=1000,\qquad z=2000
\displaystyle \text{Thus, the prices of the commodities }P,\ Q\text{ and }R\text{ are Rs }3000,\text{ Rs }1000\text{ and Rs }2000\text{ per unit, respectively.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The management committee of a residential colony decided to award some of its}
\displaystyle \text{members, say }x,\text{ for honesty, some, say }y,\text{ for helping others, and some others, say }z,\text{ for}
\displaystyle \text{supervising the workers to keep the colony neat and clean. The total number of awardees is }12.
\displaystyle \text{Three times the sum of the awardees for helping others and supervision, added to twice the}
\displaystyle \text{number of awardees for honesty, is }33.\text{ If the sum of the number of awardees for honesty}
\displaystyle \text{and supervision is twice the number of awardees for helping others, find the number of}
\displaystyle \text{awardees in each category using the matrix method. Apart from honesty, cooperation and}
\displaystyle \text{supervision, suggest one more value which the management of the colony should include}
\displaystyle \text{for awards.}\qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question,}
\displaystyle x+y+z=12
\displaystyle 2x+3(y+z)=33
\displaystyle x+z=2y
\displaystyle \therefore x+y+z=12,\qquad 2x+3y+3z=33,\qquad x-2y+z=0
\displaystyle \text{The system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\33\\0\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}12\\33\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\2&3&3\\1&-2&1\end{vmatrix}
\displaystyle =1(3+6)-1(2-3)+1(-4-3)
\displaystyle =9+1-7=3\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle A^{-1}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}12\\33\\0\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}108-99+0\\12+0+0\\-84+99+0\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}9\\12\\15\end{bmatrix}=\begin{bmatrix}3\\4\\5\end{bmatrix}
\displaystyle \therefore x=3,\qquad y=4,\qquad z=5
\displaystyle \text{Thus, the numbers of awardees for honesty, helping others and supervision are }3,\ 4\text{ and }5,\text{ respectively.}
\displaystyle \text{One more value which the management of the colony may include for awards is sincerity.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A school wants to award its students for the values of Honesty, Regularity and}
\displaystyle \text{Hard work with a total cash award of Rs }6000.\text{ Three times the award money for Hard work,}
\displaystyle \text{added to that given for Honesty, amounts to Rs }11000.\text{ The award money given for Honesty}
\displaystyle \text{and Hard work together is double the amount given for Regularity. Represent the situation}
\displaystyle \text{algebraically and find the award money for each value using the matrix method. Apart from}
\displaystyle \text{Honesty, Regularity and Hard work, suggest one more value which the school should include}
\displaystyle \text{for awards.}\qquad \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the award money given for Honesty, Regularity and Hard work be Rs }x,\text{ Rs }y\text{ and Rs }z,\text{ respectively.}
\displaystyle \text{Since the total cash award is Rs }6000,
\displaystyle x+y+z=6000\qquad (1)
\displaystyle \text{Three times the award money for Hard work, added to the award money for Honesty, is Rs }11000.
\displaystyle \therefore x+3z=11000
\displaystyle \Rightarrow x+0y+3z=11000\qquad (2)
\displaystyle \text{The award money for Honesty and Hard work together is double the amount given for Regularity.}
\displaystyle \therefore x+z=2y
\displaystyle \Rightarrow x-2y+z=0\qquad (3)
\displaystyle \text{The above system can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6000\\11000\\0\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}6000\\11000\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&0&3\\1&-2&1\end{vmatrix}=6\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle \text{adj }A=\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6&-3&3\\2&0&-2\\-2&3&-1\end{bmatrix}\begin{bmatrix}6000\\11000\\0\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}36000-33000+0\\12000+0+0\\-12000+33000+0\end{bmatrix}
\displaystyle =\frac{1}{6}\begin{bmatrix}3000\\12000\\21000\end{bmatrix}=\begin{bmatrix}500\\2000\\3500\end{bmatrix}
\displaystyle \therefore x=500,\qquad y=2000,\qquad z=3500
\displaystyle \text{Thus, the award money given for Honesty, Regularity and Hard work is Rs }500,\text{ Rs }2000\text{ and Rs }3500,\text{ respectively.}
\displaystyle \text{One more value which the school may include for awards is sincerity.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Two institutions decided to award their employees for the three values of}
\displaystyle \text{resourcefulness, competence and determination in the form of prizes at the rates of Rs }x,\ y\text{ and }z
\displaystyle \text{per person, respectively. The first institution decided to award }4,\ 3\text{ and }2\text{ employees,}
\displaystyle \text{respectively, with a total prize money of Rs }37000.\text{ The second institution decided to award}
\displaystyle \text{ }5,\ 3\text{ and }4\text{ employees, respectively, with a total prize money of Rs }47000.\text{ If all}
\displaystyle \text{three prizes per person together amount to Rs }12000,\text{ then, using the matrix method, find the}
\displaystyle \text{values of }x,\ y\text{ and }z.\text{ What values are described in these equations?}
\displaystyle \hfill \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{According to the question,}
\displaystyle 4x+3y+2z=37000
\displaystyle 5x+3y+4z=47000
\displaystyle x+y+z=12000
\displaystyle \text{The system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}37000\\47000\\12000\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}37000\\47000\\12000\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}4&3&2\\5&3&4\\1&1&1\end{vmatrix}
\displaystyle =4(3-4)-3(5-4)+2(5-3)
\displaystyle =-4-3+4=-3\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=-1,\qquad C_{12}=-1,\qquad C_{13}=2
\displaystyle C_{21}=-1,\qquad C_{22}=2,\qquad C_{23}=-1
\displaystyle C_{31}=6,\qquad C_{32}=-6,\qquad C_{33}=-3
\displaystyle \text{adj }A=\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&-1&6\\-1&2&-6\\2&-1&-3\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{3}\begin{bmatrix}-1&-1&6\\-1&2&-6\\2&-1&-3\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{3}\begin{bmatrix}-1&-1&6\\-1&2&-6\\2&-1&-3\end{bmatrix}\begin{bmatrix}37000\\47000\\12000\end{bmatrix}
\displaystyle =-\frac{1}{3}\begin{bmatrix}-37000-47000+72000\\-37000+94000-72000\\74000-47000-36000\end{bmatrix}
\displaystyle =-\frac{1}{3}\begin{bmatrix}-12000\\-15000\\-9000\end{bmatrix}=\begin{bmatrix}4000\\5000\\3000\end{bmatrix}
\displaystyle \therefore x=4000,\qquad y=5000,\qquad z=3000
\displaystyle \text{Thus, the prizes for resourcefulness, competence and determination are Rs }4000,\text{ Rs }5000\text{ and Rs }3000\text{ per person, respectively.}
\displaystyle \text{The values described in these equations are resourcefulness, competence and determination.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Two factories decided to award their employees for three values: (a) adaptability to}
\displaystyle \text{new techniques, (b) being careful and alert in difficult situations, and (c) remaining calm in}
\displaystyle \text{tense situations, at the rates of Rs }x,\text{ Rs }y\text{ and Rs }z\text{ per person, respectively. The first}
\displaystyle \text{factory decided to honour }2,\ 4\text{ and }3\text{ employees, respectively, with a total prize money}
\displaystyle \text{of Rs }29000.\text{ The second factory decided to honour }5,\ 2\text{ and }3\text{ employees, respectively,}
\displaystyle \text{with a total prize money of Rs }30500.\text{ If the three prizes per person together cost Rs }9500,\text{ then:}
\displaystyle \text{(i) represent the above situation by a matrix equation and form the linear equations using}
\displaystyle \text{matrix multiplication; (ii) solve the equations by the matrix method; (iii) which values are}
\displaystyle \text{reflected in the question?}\qquad \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the prize amounts for adaptability, alertness and calmness be Rs }x,\text{ Rs }y\text{ and Rs }z\text{ per person, respectively.}
\displaystyle \text{According to the question,}
\displaystyle 2x+4y+3z=29000\qquad (1)
\displaystyle 5x+2y+3z=30500\qquad (2)
\displaystyle x+y+z=9500\qquad (3)
\displaystyle \text{(i) The matrix equation is}
\displaystyle \begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}29000\\30500\\9500\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}29000\\30500\\9500\end{bmatrix}
\displaystyle \text{On multiplying the matrices, we obtain the linear equations}
\displaystyle 2x+4y+3z=29000,\qquad 5x+2y+3z=30500,\qquad x+y+z=9500
\displaystyle \text{(ii) Now,}
\displaystyle |A|=\begin{vmatrix}2&4&3\\5&2&3\\1&1&1\end{vmatrix}
\displaystyle =2(2-3)-4(5-3)+3(5-2)
\displaystyle =-2-8+9=-1\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=-1,\qquad C_{12}=-2,\qquad C_{13}=3
\displaystyle C_{21}=-1,\qquad C_{22}=-1,\qquad C_{23}=2
\displaystyle C_{31}=6,\qquad C_{32}=9,\qquad C_{33}=-16
\displaystyle \text{adj }A=\begin{bmatrix}-1&-2&3\\-1&-1&2\\6&9&-16\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}-1&-1&6\\-2&-1&9\\3&2&-16\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\begin{bmatrix}-1&-1&6\\-2&-1&9\\3&2&-16\end{bmatrix}
\displaystyle =\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}\begin{bmatrix}29000\\30500\\9500\end{bmatrix}
\displaystyle =\begin{bmatrix}29000+30500-57000\\58000+30500-85500\\-87000-61000+152000\end{bmatrix}
\displaystyle =\begin{bmatrix}2500\\3000\\4000\end{bmatrix}
\displaystyle \therefore x=2500,\qquad y=3000,\qquad z=4000
\displaystyle \text{Thus, the prizes for adaptability to new techniques, alertness in difficult situations and calmness in tense situations are Rs }2500,
\displaystyle \text{Rs }3000\text{ and Rs }4000\text{ per person, respectively.}
\displaystyle \text{(iii) The values reflected in the question are adaptability, carefulness, alertness and calmness under pressure.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Two schools A and B want to award their selected students for the values of}
\displaystyle \text{sincerity, truthfulness and helpfulness. School A wants to award Rs }x,\text{ Rs }y\text{ and Rs }z\text{ each}
\displaystyle \text{for the three respective values to }3,\ 2\text{ and }1\text{ students, respectively, with a total award}
\displaystyle \text{money of Rs }1600.\text{ School B wants to spend Rs }2300\text{ to award }4,\ 1\text{ and }3\text{ students}
\displaystyle \text{on the respective values by giving the same award money as before. If the total amount of one}
\displaystyle \text{prize on each value is Rs }900,\text{ then, using matrices, find the award money for each value.}
\displaystyle \text{Apart from these three values, suggest one more value which should be considered for an award.}
\displaystyle \hfill \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the award money given for sincerity, truthfulness and helpfulness be Rs }x,\text{ Rs }y\text{ and Rs }z,\text{ respectively.}
\displaystyle \text{Since the total amount of one prize on each value is Rs }900,
\displaystyle x+y+z=900\qquad (1)
\displaystyle \text{The total award money given by School A is Rs }1600.
\displaystyle 3x+2y+z=1600\qquad (2)
\displaystyle \text{The total award money given by School B is Rs }2300.
\displaystyle 4x+y+3z=2300\qquad (3)
\displaystyle \text{The above system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}
\displaystyle CX=D,\text{ where }C=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }D=\begin{bmatrix}900\\1600\\2300\end{bmatrix}
\displaystyle |C|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}
\displaystyle =1(6-1)-1(9-4)+1(3-8)
\displaystyle =5-5-5=-5\ne0
\displaystyle \therefore C^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }c_{ij}\text{ in }C=[c_{ij}].
\displaystyle C_{11}=5,\qquad C_{12}=-5,\qquad C_{13}=-5
\displaystyle C_{21}=-2,\qquad C_{22}=-1,\qquad C_{23}=3
\displaystyle C_{31}=-1,\qquad C_{32}=2,\qquad C_{33}=-1
\displaystyle \text{adj }C=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle C^{-1}=\frac{1}{|C|}\text{adj }C
\displaystyle =-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle \therefore X=C^{-1}D
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}\begin{bmatrix}900\\1600\\2300\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}4500-3200-2300\\-4500-1600+4600\\-4500+4800-2300\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}-1000\\-1500\\-2000\end{bmatrix}=\begin{bmatrix}200\\300\\400\end{bmatrix}
\displaystyle \therefore x=200,\qquad y=300,\qquad z=400
\displaystyle \text{Thus, the award money for sincerity, truthfulness and helpfulness is Rs }200,\text{ Rs }300\text{ and Rs }400,\text{ respectively.}
\displaystyle \text{One more value which should be considered for an award is hard work.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Two schools P and Q want to award their selected students for the values of}
\displaystyle \text{Discipline, Politeness and Punctuality. School P wants to award Rs }x,\text{ Rs }y\text{ and Rs }z\text{ each}
\displaystyle \text{for the three respective values to }3,\ 2\text{ and }1\text{ students, respectively, with a total award}
\displaystyle \text{money of Rs }1000.\text{ School Q wants to spend Rs }1500\text{ to award }4,\ 1\text{ and }3\text{ students}
\displaystyle \text{on the respective values by giving the same award money as before. If the total amount of one}
\displaystyle \text{prize on each value is Rs }600,\text{ then, using matrices, find the award money for each value.}
\displaystyle \text{Apart from these three values, suggest one more value for awards.}
\displaystyle \hfill \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the award money given for Discipline, Politeness and Punctuality be Rs }x,\text{ Rs }y\text{ and Rs }z,\text{ respectively.}
\displaystyle \text{Since the total amount of one prize on each value is Rs }600,
\displaystyle x+y+z=600\qquad (1)
\displaystyle \text{The total award money given by School P is Rs }1000.
\displaystyle 3x+2y+z=1000\qquad (2)
\displaystyle \text{The total award money given by School Q is Rs }1500.
\displaystyle 4x+y+3z=1500\qquad (3)
\displaystyle \text{The above system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}600\\1000\\1500\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}600\\1000\\1500\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}
\displaystyle =1(6-1)-1(9-4)+1(3-8)
\displaystyle =5-5-5=-5\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=5,\qquad C_{12}=-5,\qquad C_{13}=-5
\displaystyle C_{21}=-2,\qquad C_{22}=-1,\qquad C_{23}=3
\displaystyle C_{31}=-1,\qquad C_{32}=2,\qquad C_{33}=-1
\displaystyle \text{adj }A=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}\begin{bmatrix}600\\1000\\1500\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}3000-2000-1500\\-3000-1000+3000\\-3000+3000-1500\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}-500\\-1000\\-1500\end{bmatrix}=\begin{bmatrix}100\\200\\300\end{bmatrix}
\displaystyle \therefore x=100,\qquad y=200,\qquad z=300
\displaystyle \text{Thus, the award money for Discipline, Politeness and Punctuality is Rs }100,\text{ Rs }200\text{ and Rs }300,\text{ respectively.}
\displaystyle \text{One more value which should be considered for an award is honesty.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Two schools P and Q want to award their selected students for the values of}
\displaystyle \text{Tolerance, Kindness and Leadership. School P wants to award Rs }x,\text{ Rs }y\text{ and Rs }z\text{ each}
\displaystyle \text{for the three respective values to }3,\ 2\text{ and }1\text{ students, respectively, with a total award}
\displaystyle \text{money of Rs }2200.\text{ School Q wants to spend Rs }3100\text{ to award }4,\ 1\text{ and }3\text{ students}
\displaystyle \text{on the respective values by giving the same award money as School P. If the total amount of one}
\displaystyle \text{prize on each value is Rs }1200,\text{ then, using matrices, find the award money for each value.}
\displaystyle \text{Apart from these three values, suggest one more value which should be considered for an award.}
\displaystyle \hfill \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the award money given for Tolerance, Kindness and Leadership be Rs }x,\text{ Rs }y\text{ and Rs }z,\text{ respectively.}
\displaystyle \text{Since the total amount of one prize on each value is Rs }1200,
\displaystyle x+y+z=1200\qquad (1)
\displaystyle \text{The total award money given by School P is Rs }2200.
\displaystyle 3x+2y+z=2200\qquad (2)
\displaystyle \text{The total award money given by School Q is Rs }3100.
\displaystyle 4x+y+3z=3100\qquad (3)
\displaystyle \text{The above system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1200\\2200\\3100\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }B=\begin{bmatrix}1200\\2200\\3100\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}
\displaystyle =1(6-1)-1(9-4)+1(3-8)
\displaystyle =5-5-5=-5\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=5,\qquad C_{12}=-5,\qquad C_{13}=-5
\displaystyle C_{21}=-2,\qquad C_{22}=-1,\qquad C_{23}=3
\displaystyle C_{31}=-1,\qquad C_{32}=2,\qquad C_{33}=-1
\displaystyle \text{adj }A=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}\begin{bmatrix}1200\\2200\\3100\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}6000-4400-3100\\-6000-2200+6200\\-6000+6600-3100\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}-1500\\-2000\\-2500\end{bmatrix}=\begin{bmatrix}300\\400\\500\end{bmatrix}
\displaystyle \therefore x=300,\qquad y=400,\qquad z=500
\displaystyle \text{Thus, the award money for Tolerance, Kindness and Leadership is Rs }300,\text{ Rs }400\text{ and Rs }500,\text{ respectively.}
\displaystyle \text{One more value which should be considered for an award is honesty.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A total amount of Rs }7000\text{ is deposited in three different savings bank accounts}
\displaystyle \text{at annual interest rates of }5\%,\ 8\%\text{ and }8\frac{1}{2}\%\text{ respectively. The total annual}
\displaystyle \text{interest from these three accounts is Rs }550.\text{ Equal amounts have been deposited in the}
\displaystyle \text{ }5\%\text{ and }8\%\text{ savings accounts. Find the amount deposited in each of the three accounts}
\displaystyle \text{with the help of matrices.}\qquad \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the amounts deposited in the accounts earning }5\%,\ 8\%\text{ and }8\frac{1}{2}\%\text{ interest be Rs }x,\text{ Rs }x\text{ and Rs }y,\text{ respectively.}
\displaystyle \text{Since the total amount deposited is Rs }7000,
\displaystyle x+x+y=7000
\displaystyle \Rightarrow 2x+y=7000\qquad (1)
\displaystyle \text{The total annual interest is Rs }550.
\displaystyle \therefore \frac{5}{100}x+\frac{8}{100}x+\frac{17}{200}y=550
\displaystyle \Rightarrow 10x+16x+17y=110000
\displaystyle \Rightarrow 26x+17y=110000\qquad (2)
\displaystyle \text{The above system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&1\\26&17\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7000\\110000\end{bmatrix}
\displaystyle AX=B,\text{ where }A=\begin{bmatrix}2&1\\26&17\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}7000\\110000\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&1\\26&17\end{vmatrix}=34-26=8\ne0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }a_{ij}\text{ in }A=[a_{ij}].
\displaystyle C_{11}=17,\qquad C_{12}=-26,\qquad C_{21}=-1,\qquad C_{22}=2
\displaystyle \text{adj }A=\begin{bmatrix}17&-26\\-1&2\end{bmatrix}^{T}=\begin{bmatrix}17&-1\\-26&2\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\text{adj }A
\displaystyle =\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}
\displaystyle \therefore X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{8}\begin{bmatrix}17&-1\\-26&2\end{bmatrix}\begin{bmatrix}7000\\110000\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}119000-110000\\-182000+220000\end{bmatrix}
\displaystyle =\frac{1}{8}\begin{bmatrix}9000\\38000\end{bmatrix}=\begin{bmatrix}1125\\4750\end{bmatrix}
\displaystyle \therefore x=1125,\qquad y=4750
\displaystyle \text{Thus, the amounts deposited in the accounts earning }5\%,\ 8\%\text{ and }8\frac{1}{2}\%\text{ interest are Rs }1125,\text{ Rs }1125\text{ and Rs }4750,\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A shopkeeper has three varieties of pens A, B and C. Meenu purchased one pen}
\displaystyle \text{of each variety for a total of Rs }21.\text{ Jeevan purchased }4\text{ pens of variety A, }3\text{ pens}
\displaystyle \text{of variety B and }2\text{ pens of variety C for Rs }60.\text{ Shikha purchased }6\text{ pens of variety A,}
\displaystyle \text{ }2\text{ pens of variety B and }3\text{ pens of variety C for Rs }70.\text{ Using the matrix method,}
\displaystyle \text{find the cost of each variety of pen.}\qquad \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the costs of one pen of varieties A, B and C be Rs }x,\text{ Rs }y\text{ and Rs }z,\text{ respectively.}
\displaystyle \text{Meenu purchased one pen of each variety for Rs }21.
\displaystyle \therefore x+y+z=21\qquad (1)
\displaystyle \text{Jeevan purchased }4\text{ pens of variety A, }3\text{ pens of variety B and }2\text{ pens of variety C for Rs }60.
\displaystyle \therefore 4x+3y+2z=60\qquad (2)
\displaystyle \text{Shikha purchased }6\text{ pens of variety A, }2\text{ pens of variety B and }3\text{ pens of variety C for Rs }70.
\displaystyle \therefore 6x+2y+3z=70\qquad (3)
\displaystyle \text{The above system of equations can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}21\\60\\70\end{bmatrix}
\displaystyle PX=Q,\text{ where }P=\begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }Q=\begin{bmatrix}21\\60\\70\end{bmatrix}
\displaystyle |P|=\begin{vmatrix}1&1&1\\4&3&2\\6&2&3\end{vmatrix}
\displaystyle =1(9-4)-1(12-12)+1(8-18)
\displaystyle =5-0-10=-5\ne0
\displaystyle \therefore P^{-1}\text{ exists.}
\displaystyle \text{Let }C_{ij}\text{ be the cofactors of the elements }p_{ij}\text{ in }P=[p_{ij}].
\displaystyle C_{11}=5,\qquad C_{12}=0,\qquad C_{13}=-10
\displaystyle C_{21}=-1,\qquad C_{22}=-3,\qquad C_{23}=4
\displaystyle C_{31}=-1,\qquad C_{32}=2,\qquad C_{33}=-1
\displaystyle \text{adj }P=\begin{bmatrix}5&0&-10\\-1&-3&4\\-1&2&-1\end{bmatrix}^{T}
\displaystyle =\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}
\displaystyle P^{-1}=\frac{1}{|P|}\text{adj }P
\displaystyle =-\frac{1}{5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}
\displaystyle \therefore X=P^{-1}Q
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}\begin{bmatrix}21\\60\\70\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}105-60-70\\-180+140\\-210+240-70\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}-25\\-40\\-40\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}
\displaystyle \therefore x=5,\qquad y=8,\qquad z=8
\displaystyle \text{Thus, the costs of one pen of varieties A, B and C are Rs }5,\text{ Rs }8\text{ and Rs }8,\text{ respectively.}
\displaystyle \\

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