\displaystyle \textbf{Question 1: }\ \ \text{Solve the following system of equations, using matrix method:}
\displaystyle x+2y+z=7,\ \ x+3z=11,\ \ 2x-3y=1\qquad [\text{CBSE 2002, 2003, 2005}]

\displaystyle \text{Answer:}
\displaystyle \text{The given system of equations is}
\displaystyle x+2y+z=7
\displaystyle x+0y+3z=11
\displaystyle 2x-3y+0z=1

\displaystyle \text{or,}\quad \begin{bmatrix} 1&2&1\\ 1&0&3\\ 2&-3&0 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 7\\ 11\\ 1 \end{bmatrix}

\displaystyle \text{or,}\quad AX=B,\ \text{where }A=\begin{bmatrix} 1&2&1\\ 1&0&3\\ 2&-3&0 \end{bmatrix},\ X=\begin{bmatrix} x\\ y\\ z \end{bmatrix}

\displaystyle \text{and }B=\begin{bmatrix} 7\\ 11\\ 1 \end{bmatrix}

\displaystyle \text{Now,}\quad |A|=\begin{vmatrix} 1&2&1\\ 1&0&3\\ 2&-3&0 \end{vmatrix}=1(0+9)-2(0-6)+1(-3-0)

\displaystyle =9+12-3=18\neq 0

\displaystyle \text{So, the given system of equations has a unique solution given by }X=A^{-1}B.

\displaystyle \text{Let }C_{ij}\text{ be the co-factors of elements }a_{ij}\text{ in }A=[a_{ij}].\text{ Then,}

\displaystyle C_{11}=(-1)^{1+1}\begin{vmatrix} 0&3\\ -3&0 \end{vmatrix}=9,\quad C_{12}=(-1)^{1+2}\begin{vmatrix} 1&3\\ 2&0 \end{vmatrix}=6,

\displaystyle C_{13}=(-1)^{1+3}\begin{vmatrix} 1&0\\ 2&-3 \end{vmatrix}=-3,\quad C_{21}=(-1)^{2+1}\begin{vmatrix} 2&1\\ -3&0 \end{vmatrix}=-3,

\displaystyle C_{22}=(-1)^{2+2}\begin{vmatrix} 1&1\\ 2&0 \end{vmatrix}=-2,\quad C_{23}=(-1)^{2+3}\begin{vmatrix} 1&2\\ 2&-3 \end{vmatrix}=7,

\displaystyle C_{31}=(-1)^{3+1}\begin{vmatrix} 2&1\\ 0&3 \end{vmatrix}=6,\quad C_{32}=(-1)^{3+2}\begin{vmatrix} 1&1\\ 1&3 \end{vmatrix}=-2,

\displaystyle \text{and,}\quad C_{33}=(-1)^{3+3}\begin{vmatrix} 1&2\\ 1&0 \end{vmatrix}=-2

\displaystyle \therefore\ \mathrm{adj}\ A=\begin{bmatrix} 9&6&-3\\ -3&-2&7\\ 6&-2&-2 \end{bmatrix}^{T}=\begin{bmatrix} 9&-3&6\\ 6&-2&-2\\ -3&7&-2 \end{bmatrix}

\displaystyle \Rightarrow\ A^{-1}=\frac{1}{|A|}\mathrm{adj}\ A=\frac{1}{18}\begin{bmatrix} 9&-3&6\\ 6&-2&-2\\ -3&7&-2 \end{bmatrix}

\displaystyle \text{Now,}\quad X=A^{-1}B

\displaystyle \Rightarrow\ X=\frac{1}{18}\begin{bmatrix} 9&-3&6\\ 6&-2&-2\\ -3&7&-2 \end{bmatrix}\begin{bmatrix} 7\\ 11\\ 1 \end{bmatrix}

\displaystyle =\frac{1}{18}\begin{bmatrix} 63-33+6\\ 42-22-2\\ -21+77-2 \end{bmatrix}=\frac{1}{18}\begin{bmatrix} 36\\ 18\\ 54 \end{bmatrix}=\begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix}

\displaystyle \Rightarrow\ \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix}\ \Rightarrow\ x=2,\ y=1\ \text{and}\ z=3

\displaystyle \text{Hence, }x=2,\ y=1\ \text{and}\ z=3\text{ is the required solution.}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.