\displaystyle \textbf{Question 1: } \text{If a matrix has }8\text{ elements, what are the possible orders it can have?}
\displaystyle \text{What if it has }5\text{ elements?}
\displaystyle \text{Answer:}
\displaystyle \text{If a matrix is of order }m\times n,\text{ then it has }mn\text{ elements.}
\displaystyle \text{If the matrix has }8\text{ elements, then }mn=8.
\displaystyle \text{The possible ordered pairs }(m,n)\text{ are }(1,8),(8,1),(2,4)\text{ and }(4,2).
\displaystyle \therefore \text{The possible orders are }1\times8,\ 8\times1,\ 2\times4\text{ and }4\times2.
\displaystyle \text{If the matrix has }5\text{ elements, then }mn=5.
\displaystyle \text{The possible ordered pairs }(m,n)\text{ are }(1,5)\text{ and }(5,1).
\displaystyle \therefore \text{The possible orders are }1\times5\text{ and }5\times1.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{If }A=[a_{ij}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}\text{ and }
\displaystyle B=[b_{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&2\end{bmatrix},\text{ then find:}
\displaystyle \text{(i) }a_{22}+b_{21}
\displaystyle \text{(ii) }a_{11}\times b_{11}+a_{22}\times b_{22}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=[a_{ij}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}\text{ and }B=[b_{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&2\end{bmatrix}.
\displaystyle \text{(i)}
\displaystyle a_{22}+b_{21}=4+(-3)=1.
\displaystyle \text{(ii)}
\displaystyle a_{11}\times b_{11}+a_{22}\times b_{22}=2\times2+4\times4=4+16=20.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Let }A\text{ be a matrix of order }3\times4.\text{ If }R_1\text{ denotes the first row of }A
\displaystyle \text{and }C_2\text{ denotes its second column, then determine the order of matrices }R_1\text{ and }C_2.
\displaystyle \text{Answer:}
\displaystyle A\text{ is a matrix of order }3\times4.
\displaystyle \therefore A=\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{bmatrix}.
\displaystyle \text{The first row is }R_1=\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\end{bmatrix}.
\displaystyle \text{It has }1\text{ row and }4\text{ columns.}
\displaystyle \therefore \text{The order of }R_1\text{ is }1\times4.
\displaystyle \text{The second column is }C_2=\begin{bmatrix}a_{12}\\a_{22}\\a_{32}\end{bmatrix}.
\displaystyle \text{It has }3\text{ rows and }1\text{ column.}
\displaystyle \therefore \text{The order of }C_2\text{ is }3\times1.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Construct a }2\times3\text{ matrix }A=[a_{ij}]\text{ whose elements }a_{ij}\text{ are given by}
\displaystyle \text{(i) }a_{ij}=i\times j \qquad \text{(ii) }a_{ij}=2i-j \qquad \text{(iii) }a_{ij}=i+j \qquad \text{(iv) }a_{ij}=\frac{(i+j)^2}{2}
\displaystyle \text{Answer:}
\displaystyle \text{For a }2\times3\text{ matrix, }1\le i\le2\text{ and }1\le j\le3.
\displaystyle \therefore A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}.

\displaystyle \text{(i)}
\displaystyle a_{11}=1,\ a_{12}=2,\ a_{13}=3,\ a_{21}=2,\ a_{22}=4,\ a_{23}=6.
\displaystyle \therefore A=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}.

\displaystyle \text{(ii)}
\displaystyle a_{11}=2(1)-1=1,\ a_{12}=2(1)-2=0,\ a_{13}=2(1)-3=-1.
\displaystyle a_{21}=2(2)-1=3,\ a_{22}=2(2)-2=2,\ a_{23}=2(2)-3=1.
\displaystyle \therefore A=\begin{bmatrix}1&0&-1\\3&2&1\end{bmatrix}.

\displaystyle \text{(iii)}
\displaystyle a_{11}=1+1=2,\ a_{12}=1+2=3,\ a_{13}=1+3=4.
\displaystyle a_{21}=2+1=3,\ a_{22}=2+2=4,\ a_{23}=2+3=5.
\displaystyle \therefore A=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}.

\displaystyle \text{(iv)}
\displaystyle a_{11}=\frac{(1+1)^2}{2}=2,\ a_{12}=\frac{(1+2)^2}{2}=\frac92,\ a_{13}=\frac{(1+3)^2}{2}=8.
\displaystyle a_{21}=\frac{(2+1)^2}{2}=\frac92,\ a_{22}=\frac{(2+2)^2}{2}=8,\ a_{23}=\frac{(2+3)^2}{2}=\frac{25}{2}.
\displaystyle \therefore A=\begin{bmatrix}2&\frac92&8\\\frac92&8&\frac{25}{2}\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Construct a }2\times2\text{ matrix }A=[a_{ij}]\text{ whose elements }a_{ij}\text{ are given by}
\displaystyle \text{(i) }a_{ij}=\frac{(i+j)^2}{2}\qquad \text{(ii) }a_{ij}=\frac{(i-j)^2}{2}\qquad \text{(iii) }a_{ij}=\frac{(i-2j)^2}{2}
\displaystyle \text{(iv) }a_{ij}=\frac{(2i+j)^2}{2}\hfill[\text{CBSE 2002}]
\displaystyle \text{(v) }a_{ij}=\frac{|2i-3j|}{2}\qquad \text{(vi) }a_{ij}=\frac{|-3i+j|}{2}\qquad \text{(vii) }a_{ij}=e^{2ix}\sin(xj)
\displaystyle \text{Answer:}
\displaystyle \text{For a }2\times2\text{ matrix, }1\le i,j\le2.
\displaystyle \therefore A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}.

\displaystyle \text{(i)}
\displaystyle a_{11}=2,\ a_{12}=\frac92,\ a_{21}=\frac92,\ a_{22}=8.
\displaystyle \therefore A=\begin{bmatrix}2&\frac92\\\frac92&8\end{bmatrix}.

\displaystyle \text{(ii)}
\displaystyle a_{11}=0,\ a_{12}=\frac12,\ a_{21}=\frac12,\ a_{22}=0.
\displaystyle \therefore A=\begin{bmatrix}0&\frac12\\\frac12&0\end{bmatrix}.

\displaystyle \text{(iii)}
\displaystyle a_{11}=\frac12,\ a_{12}=\frac92,\ a_{21}=0,\ a_{22}=2.
\displaystyle \therefore A=\begin{bmatrix}\frac12&\frac92\\0&2\end{bmatrix}.

\displaystyle \text{(iv)}
\displaystyle a_{11}=\frac92,\ a_{12}=8,\ a_{21}=\frac{25}{2},\ a_{22}=18.
\displaystyle \therefore A=\begin{bmatrix}\frac92&8\\\frac{25}{2}&18\end{bmatrix}.

\displaystyle \text{(v)}
\displaystyle a_{11}=\frac12,\ a_{12}=2,\ a_{21}=\frac12,\ a_{22}=1.
\displaystyle \therefore A=\begin{bmatrix}\frac12&2\\\frac12&1\end{bmatrix}.

\displaystyle \text{(vi)}
\displaystyle a_{11}=1,\ a_{12}=\frac12,\ a_{21}=\frac52,\ a_{22}=2.
\displaystyle \therefore A=\begin{bmatrix}1&\frac12\\\frac52&2\end{bmatrix}.

\displaystyle \text{(vii)}
\displaystyle a_{11}=e^{2x}\sin x,\ a_{12}=e^{2x}\sin2x,\ a_{21}=e^{4x}\sin x,\ a_{22}=e^{4x}\sin2x.
\displaystyle \therefore A=\begin{bmatrix}e^{2x}\sin x&e^{2x}\sin2x\\e^{4x}\sin x&e^{4x}\sin2x\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Construct a }3\times4\text{ matrix }A=[a_{ij}]\text{ whose elements }a_{ij}\text{ are}
\displaystyle \text{given by:}
\displaystyle \text{(i) }a_{ij}=i+j\qquad\text{(ii) }a_{ij}=i-j\qquad\text{(iii) }a_{ij}=2i
\displaystyle \text{(iv) }a_{ij}=j\qquad\text{(v) }a_{ij}=\frac{1}{2}|-3i+j|
\displaystyle \text{Answer:}
\displaystyle \text{For a }3\times4\text{ matrix, }1\le i\le3\text{ and }1\le j\le4.
\displaystyle \therefore A=\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{bmatrix}.

\displaystyle \text{(i) When }a_{ij}=i+j,
\displaystyle a_{11}=2,\ a_{12}=3,\ a_{13}=4,\ a_{14}=5.
\displaystyle a_{21}=3,\ a_{22}=4,\ a_{23}=5,\ a_{24}=6.
\displaystyle a_{31}=4,\ a_{32}=5,\ a_{33}=6,\ a_{34}=7.
\displaystyle \therefore A=\begin{bmatrix}2&3&4&5\\3&4&5&6\\4&5&6&7\end{bmatrix}.

\displaystyle \text{(ii) When }a_{ij}=i-j,
\displaystyle a_{11}=0,\ a_{12}=-1,\ a_{13}=-2,\ a_{14}=-3.
\displaystyle a_{21}=1,\ a_{22}=0,\ a_{23}=-1,\ a_{24}=-2.
\displaystyle a_{31}=2,\ a_{32}=1,\ a_{33}=0,\ a_{34}=-1.
\displaystyle \therefore A=\begin{bmatrix}0&-1&-2&-3\\1&0&-1&-2\\2&1&0&-1\end{bmatrix}.

\displaystyle \text{(iii) When }a_{ij}=2i,
\displaystyle a_{11}=a_{12}=a_{13}=a_{14}=2.
\displaystyle a_{21}=a_{22}=a_{23}=a_{24}=4.
\displaystyle a_{31}=a_{32}=a_{33}=a_{34}=6.
\displaystyle \therefore A=\begin{bmatrix}2&2&2&2\\4&4&4&4\\6&6&6&6\end{bmatrix}.

\displaystyle \text{(iv) When }a_{ij}=j,
\displaystyle a_{11}=1,\ a_{12}=2,\ a_{13}=3,\ a_{14}=4.
\displaystyle a_{21}=1,\ a_{22}=2,\ a_{23}=3,\ a_{24}=4.
\displaystyle a_{31}=1,\ a_{32}=2,\ a_{33}=3,\ a_{34}=4.
\displaystyle \therefore A=\begin{bmatrix}1&2&3&4\\1&2&3&4\\1&2&3&4\end{bmatrix}.

\displaystyle \text{(v) When }a_{ij}=\frac{1}{2}|-3i+j|,
\displaystyle a_{11}=1,\ a_{12}=\frac12,\ a_{13}=0,\ a_{14}=\frac12.
\displaystyle a_{21}=\frac52,\ a_{22}=2,\ a_{23}=\frac32,\ a_{24}=1.
\displaystyle a_{31}=4,\ a_{32}=\frac72,\ a_{33}=3,\ a_{34}=\frac52.
\displaystyle \therefore A=\begin{bmatrix}1&\frac12&0&\frac12\\\frac52&2&\frac32&1\\4&\frac72&3&\frac52\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Construct a }4\times3\text{ matrix }A=[a_{ij}]\text{ whose elements }a_{ij}\text{ are given by:}
\displaystyle \text{(i) }a_{ij}=2i+\frac{i}{j}\qquad\text{(ii) }a_{ij}=\frac{i-j}{i+j}\qquad\text{(iii) }a_{ij}=i
\displaystyle \text{Answer:}
\displaystyle \text{For a }4\times3\text{ matrix, }1\le i\le4\text{ and }1\le j\le3.
\displaystyle \therefore A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\end{bmatrix}.

\displaystyle \text{(i) When }a_{ij}=2i+\frac{i}{j},
\displaystyle a_{11}=2(1)+\frac11=3,\quad a_{12}=2(1)+\frac12=\frac52,\quad a_{13}=2(1)+\frac13=\frac73.
\displaystyle a_{21}=2(2)+\frac21=6,\quad a_{22}=2(2)+\frac22=5,\quad a_{23}=2(2)+\frac23=\frac{14}{3}.
\displaystyle a_{31}=2(3)+\frac31=9,\quad a_{32}=2(3)+\frac32=\frac{15}{2},\quad a_{33}=2(3)+\frac33=7.
\displaystyle a_{41}=2(4)+\frac41=12,\quad a_{42}=2(4)+\frac42=10,\quad a_{43}=2(4)+\frac43=\frac{28}{3}.
\displaystyle \therefore A=\begin{bmatrix}3&\frac52&\frac73\\6&5&\frac{14}{3}\\9&\frac{15}{2}&7\\12&10&\frac{28}{3}\end{bmatrix}.

\displaystyle \text{(ii) When }a_{ij}=\frac{i-j}{i+j},
\displaystyle a_{11}=0,\quad a_{12}=-\frac13,\quad a_{13}=-\frac12.
\displaystyle a_{21}=\frac13,\quad a_{22}=0,\quad a_{23}=-\frac15.
\displaystyle a_{31}=\frac12,\quad a_{32}=\frac15,\quad a_{33}=0.
\displaystyle a_{41}=\frac35,\quad a_{42}=\frac13,\quad a_{43}=\frac17.
\displaystyle \therefore A=\begin{bmatrix}0&-\frac13&-\frac12\\\frac13&0&-\frac15\\\frac12&\frac15&0\\\frac35&\frac13&\frac17\end{bmatrix}.

\displaystyle \text{(iii) When }a_{ij}=i,
\displaystyle a_{11}=a_{12}=a_{13}=1.
\displaystyle a_{21}=a_{22}=a_{23}=2.
\displaystyle a_{31}=a_{32}=a_{33}=3.
\displaystyle a_{41}=a_{42}=a_{43}=4.
\displaystyle \therefore A=\begin{bmatrix}1&1&1\\2&2&2\\3&3&3\\4&4&4\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{Find }x,\ y,\ a\text{ and }b\text{ if}
\displaystyle \begin{bmatrix}3x+4y&2&x-2y\\a+b&2a-b&-1\end{bmatrix}=\begin{bmatrix}2&2&4\\5&-5&-1\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle 3x+4y=2\qquad\cdots(1)
\displaystyle x-2y=4\qquad\cdots(2)
\displaystyle a+b=5\qquad\cdots(3)
\displaystyle 2a-b=-5\qquad\cdots(4)
\displaystyle \text{Multiplying equation (2) by }2\text{ and adding it to equation (1),}
\displaystyle 3x+4y+2x-4y=2+8
\displaystyle 5x=10
\displaystyle \therefore x=2
\displaystyle \text{Substituting }x=2\text{ in equation (1),}
\displaystyle 3(2)+4y=2
\displaystyle 6+4y=2
\displaystyle 4y=-4
\displaystyle \therefore y=-1
\displaystyle \text{Adding equations (3) and (4),}
\displaystyle a+b+2a-b=5+(-5)
\displaystyle 3a=0
\displaystyle \therefore a=0
\displaystyle \text{Substituting }a=0\text{ in equation (3),}
\displaystyle b=5
\displaystyle \therefore x=2,\ y=-1,\ a=0,\ b=5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find }x,\ y,\ a\text{ and }b\text{ if}
\displaystyle \begin{bmatrix}2x-3y&a-b&3\\1&x+4&3a+4b\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle 2x-3y=1\qquad\cdots(1)
\displaystyle x+4=6\qquad\cdots(2)
\displaystyle \text{From equation (2),}
\displaystyle x=2
\displaystyle \text{Substituting }x=2\text{ in equation (1),}
\displaystyle 2(2)-3y=1
\displaystyle 4-3y=1
\displaystyle -3y=-3
\displaystyle \therefore y=1
\displaystyle a-b=-2\qquad\cdots(3)
\displaystyle 3a+4b=29\qquad\cdots(4)
\displaystyle \text{From equation (3),}
\displaystyle a=b-2
\displaystyle \text{Substituting }a=b-2\text{ in equation (4),}
\displaystyle 3(b-2)+4b=29
\displaystyle 3b-6+4b=29
\displaystyle 7b=35
\displaystyle \therefore b=5
\displaystyle \therefore a=5-2=3
\displaystyle \therefore x=2,\ y=1,\ a=3,\ b=5.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the values of }a,\ b,\ c\text{ and }d\text{ from the following equation:}
\displaystyle \begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle 2a+b=4\qquad\cdots(1)
\displaystyle a-2b=-3\qquad\cdots(2)
\displaystyle 5c-d=11\qquad\cdots(3)
\displaystyle 4c+3d=24\qquad\cdots(4)
\displaystyle \text{Multiplying equation (1) by }2\text{ and adding it to equation (2),}
\displaystyle 4a+2b+a-2b=8-3
\displaystyle 5a=5
\displaystyle \therefore a=1
\displaystyle \text{Substituting }a=1\text{ in equation (1),}
\displaystyle 2(1)+b=4
\displaystyle b=2
\displaystyle \text{Multiplying equation (3) by }3\text{ and adding it to equation (4),}
\displaystyle 15c-3d+4c+3d=33+24
\displaystyle 19c=57
\displaystyle \therefore c=3
\displaystyle \text{Substituting }c=3\text{ in equation (4),}
\displaystyle 4(3)+3d=24
\displaystyle 12+3d=24
\displaystyle 3d=12
\displaystyle \therefore d=4
\displaystyle \therefore a=1,\ b=2,\ c=3,\ d=4.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find }x,\ y\text{ and }z\text{ so that }A=B,\text{ where}
\displaystyle A=\begin{bmatrix}x-2&3&2z\\18z&y+2&6z\end{bmatrix},\quad B=\begin{bmatrix}y&z&6\\6y&x&2y\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x-2=y\qquad\cdots(1)
\displaystyle 3=z\qquad\cdots(2)
\displaystyle 2z=6\qquad\cdots(3)
\displaystyle 18z=6y\qquad\cdots(4)
\displaystyle y+2=x\qquad\cdots(5)
\displaystyle 6z=2y\qquad\cdots(6)
\displaystyle \text{From equation (2),}
\displaystyle z=3
\displaystyle \text{Substituting }z=3\text{ in equation (6),}
\displaystyle 6(3)=2y
\displaystyle 18=2y
\displaystyle \therefore y=9
\displaystyle \text{Substituting }y=9\text{ in equation (1),}
\displaystyle x-2=9
\displaystyle \therefore x=11
\displaystyle \text{Also, }2z=2(3)=6,\quad18z=18(3)=54=6(9)=6y
\displaystyle \text{and }y+2=9+2=11=x.
\displaystyle \therefore x=11,\ y=9,\ z=3.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\begin{bmatrix}x&3x-y\\2x+z&3y-\omega\end{bmatrix}=\begin{bmatrix}3&2\\4&7\end{bmatrix},
\displaystyle \text{find }x,\ y,\ z\text{ and }\omega.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x=3\qquad\cdots(1)
\displaystyle 3x-y=2\qquad\cdots(2)
\displaystyle 2x+z=4\qquad\cdots(3)
\displaystyle 3y-\omega=7\qquad\cdots(4)
\displaystyle \text{Substituting }x=3\text{ in equation (2),}
\displaystyle 3(3)-y=2
\displaystyle 9-y=2
\displaystyle \therefore y=7
\displaystyle \text{Substituting }y=7\text{ in equation (4),}
\displaystyle 3(7)-\omega=7
\displaystyle 21-\omega=7
\displaystyle \therefore \omega=14
\displaystyle \text{Substituting }x=3\text{ in equation (3),}
\displaystyle 2(3)+z=4
\displaystyle 6+z=4
\displaystyle \therefore z=-2
\displaystyle \therefore x=3,\ y=7,\ z=-2,\ \omega=14.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\begin{bmatrix}x-y&z\\2x-y&\omega\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix},
\displaystyle \text{find }x,\ y,\ z\text{ and }\omega.\hspace{3.5cm}[\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x-y=-1\qquad\cdots(1)
\displaystyle z=4
\displaystyle 2x-y=0\qquad\cdots(2)
\displaystyle \omega=5
\displaystyle \text{From equation (1),}
\displaystyle x=y-1
\displaystyle \text{Substituting }x=y-1\text{ in equation (2),}
\displaystyle 2(y-1)-y=0
\displaystyle 2y-2-y=0
\displaystyle y-2=0
\displaystyle \therefore y=2
\displaystyle \therefore x=2-1=1
\displaystyle \therefore x=1,\ y=2,\ z=4,\ \omega=5.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If}
\displaystyle \begin{bmatrix}x+3&z+4&2y-7\\4x+6&a-1&0\\b-3&3b&z+2c\end{bmatrix}=\begin{bmatrix}0&6&3y-2\\2x&-3&2c+2\\2b+4&-21&0\end{bmatrix},
\displaystyle \text{obtain the values of }a,\ b,\ c,\ x,\ y\text{ and }z.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x+3=0
\displaystyle \therefore x=-3
\displaystyle z+4=6
\displaystyle \therefore z=2
\displaystyle 2y-7=3y-2
\displaystyle 2y-3y=-2+7
\displaystyle -y=5
\displaystyle \therefore y=-5
\displaystyle 4x+6=2x
\displaystyle 2x=-6
\displaystyle \therefore x=-3
\displaystyle a-1=-3
\displaystyle \therefore a=-2
\displaystyle 0=2c+2
\displaystyle 2c=-2
\displaystyle \therefore c=-1
\displaystyle b-3=2b+4
\displaystyle b-2b=4+3
\displaystyle -b=7
\displaystyle \therefore b=-7
\displaystyle 3b=-21
\displaystyle \therefore b=-7
\displaystyle z+2c=0
\displaystyle 2+2(-1)=0
\displaystyle \therefore a=-2,\ b=-7,\ c=-1,\ x=-3,\ y=-5,\ z=2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\begin{bmatrix}2x+1&5x\\0&y^2+1\end{bmatrix}=\begin{bmatrix}x+3&10\\0&26\end{bmatrix},\text{ find the value(s) of }(x+y).\hspace{0.5cm}[\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle 2x+1=x+3\qquad\cdots(1)
\displaystyle 5x=10\qquad\cdots(2)
\displaystyle y^2+1=26\qquad\cdots(3)
\displaystyle \text{From equation (1),}
\displaystyle 2x-x=3-1
\displaystyle \therefore x=2
\displaystyle \text{Equation (2) is also satisfied since }5(2)=10.
\displaystyle \text{From equation (3),}
\displaystyle y^2=26-1
\displaystyle y^2=25
\displaystyle \therefore y=5\text{ or }-5
\displaystyle \therefore x+y=2+5=7
\displaystyle \text{or }x+y=2-5=-3
\displaystyle \therefore x+y=7\text{ or }-3.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\begin{bmatrix}xy&4\\z+6&x+y\end{bmatrix}=\begin{bmatrix}8&\omega\\0&6\end{bmatrix},
\displaystyle \text{find the values of }x,\ y,\ z\text{ and }\omega.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle xy=8\qquad\cdots(1)
\displaystyle \omega=4
\displaystyle z+6=0
\displaystyle \therefore z=-6
\displaystyle x+y=6\qquad\cdots(2)
\displaystyle \text{From equation (2), }y=6-x.
\displaystyle \text{Substituting }y=6-x\text{ in equation (1),}
\displaystyle x(6-x)=8
\displaystyle 6x-x^2=8
\displaystyle x^2-6x+8=0
\displaystyle (x-4)(x-2)=0
\displaystyle \therefore x=4\text{ or }x=2
\displaystyle \text{When }x=4,\quad y=6-4=2.
\displaystyle \text{When }x=2,\quad y=6-2=4.
\displaystyle \therefore x=4,\ y=2,\ z=-6,\ \omega=4
\displaystyle \text{or }x=2,\ y=4,\ z=-6,\ \omega=4.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Give an example of}
\displaystyle \text{(i) a row matrix which is also a column matrix}
\displaystyle \text{(ii) a diagonal matrix which is not scalar}
\displaystyle \text{(iii) a triangular matrix}
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \text{A matrix which is both a row matrix and a column matrix must be of order }1\times1.
\displaystyle \text{One such matrix is }A=\begin{bmatrix}5\end{bmatrix}.

\displaystyle \text{(ii)}
\displaystyle \text{A diagonal matrix has }a_{ij}=0\text{ for }i\ne j.
\displaystyle \text{A diagonal matrix is not scalar if its diagonal entries are not all equal.}
\displaystyle \text{One such matrix is }A=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}.

\displaystyle \text{(iii)}
\displaystyle \text{An upper triangular matrix has }a_{ij}=0\text{ for }i>j.
\displaystyle \text{One such matrix is }A=\begin{bmatrix}1&2&3\\0&4&5\\0&0&6\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The sales figures of two car dealers during January 2013 showed that}
\displaystyle \text{Dealer A sold }5\text{ deluxe, }3\text{ premium and }4\text{ standard cars, while Dealer B sold }7\text{ deluxe,}
\displaystyle \text{ }2\text{ premium and }3\text{ standard cars. Total sales over the two-month period of}
\displaystyle \text{January-February revealed that Dealer A sold }8\text{ deluxe, }7\text{ premium and }6\text{ standard cars,}
\displaystyle \text{while Dealer B sold }10\text{ deluxe, }5\text{ premium and }7\text{ standard cars.}
\displaystyle \text{Write }2\times3\text{ matrices summarizing the sales data for January and the}
\displaystyle \text{two-month period for each dealer.}
\displaystyle \text{Answer:}
\displaystyle \text{Taking the rows to represent Dealers A and B, and the columns to represent Deluxe, Premium and Standard cars,}
\displaystyle \text{the sales matrix for January is}
\displaystyle \begin{bmatrix}5&3&4\\7&2&3\end{bmatrix}
\displaystyle \text{The sales matrix for the two-month period (January-February) is}
\displaystyle \begin{bmatrix}8&7&6\\10&5&7\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{For what values of }x\text{ and }y\text{ are the following matrices equal?}
\displaystyle A=\begin{bmatrix}2x+1&2y\\0&y^2-5y\end{bmatrix},\quad B=\begin{bmatrix}x+3&y^2+2\\0&-6\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle 2x+1=x+3
\displaystyle \Rightarrow 2x-x=3-1
\displaystyle \Rightarrow x=2
\displaystyle 2y=y^2+2
\displaystyle \Rightarrow y^2-2y+2=0
\displaystyle \Rightarrow y=\frac{2\pm\sqrt{4-8}}{2}
\displaystyle \Rightarrow y=1\pm i
\displaystyle \therefore \text{There is no real value of }y\text{ satisfying this equation.}
\displaystyle y^2-5y=-6
\displaystyle \Rightarrow y^2-5y+6=0
\displaystyle \Rightarrow (y-3)(y-2)=0
\displaystyle \Rightarrow y=3\text{ or }2
\displaystyle \therefore \text{There is no real value of }y\text{ satisfying both equations.}
\displaystyle \therefore \text{The matrices }A\text{ and }B\text{ cannot be equal for any real values of }x\text{ and }y.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the values of }x\text{ and }y\text{ if}
\displaystyle \begin{bmatrix}x+10&y^2+2y\\0&-4\end{bmatrix}=\begin{bmatrix}3x+4&3\\0&y^2-5y\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle x+10=3x+4
\displaystyle \Rightarrow x-3x=4-10
\displaystyle \Rightarrow -2x=-6
\displaystyle \therefore x=3
\displaystyle y^2+2y=3
\displaystyle \Rightarrow y^2+2y-3=0
\displaystyle \Rightarrow y^2+3y-y-3=0
\displaystyle \Rightarrow y(y+3)-1(y+3)=0
\displaystyle \Rightarrow (y+3)(y-1)=0
\displaystyle \Rightarrow y=-3\text{ or }1\qquad\cdots(1)
\displaystyle y^2-5y=-4
\displaystyle \Rightarrow y^2-5y+4=0
\displaystyle \Rightarrow y^2-4y-y+4=0
\displaystyle \Rightarrow y(y-4)-1(y-4)=0
\displaystyle \Rightarrow (y-4)(y-1)=0
\displaystyle \Rightarrow y=4\text{ or }1\qquad\cdots(2)
\displaystyle \text{The common value from equations (1) and (2) is }y=1.
\displaystyle \therefore x=3,\ y=1.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the values of }a\text{ and }b\text{ if }A=B,\text{ where}
\displaystyle A=\begin{bmatrix}a+4&3b\\8&-6\end{bmatrix},\quad B=\begin{bmatrix}2a+2&b^2+2\\8&b^2-10\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the two matrices are equal, their corresponding elements are equal.}
\displaystyle a+4=2a+2
\displaystyle \Rightarrow a-2a=2-4
\displaystyle \Rightarrow -a=-2
\displaystyle \therefore a=2
\displaystyle 3b=b^2+2
\displaystyle \Rightarrow b^2-3b+2=0
\displaystyle \Rightarrow b^2-2b-b+2=0
\displaystyle \Rightarrow b(b-2)-1(b-2)=0
\displaystyle \Rightarrow (b-2)(b-1)=0
\displaystyle \Rightarrow b=2\text{ or }1\qquad\cdots(1)
\displaystyle -6=b^2-10
\displaystyle \Rightarrow b^2=4
\displaystyle \Rightarrow b=\pm2\qquad\cdots(2)
\displaystyle \text{The common value from equations (1) and (2) is }b=2.
\displaystyle \therefore a=2,\ b=2.
\displaystyle \\

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