Question

Solve the system of linear equations by matrix method:
$2x-3y+5z=11,3x+2y-4y=-5,x+y-2z=-3$

Answer 1

Given system of linear equations are

$2x-3y+5z=11$
$3x+2y-4z=-5$
$x+y-2z=-3$.

Represent it in matrix form

$\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$ which is in the form of $AX=B$

$A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$

$|A|=0-6+5=-1\ne 0$

$\therefore A^{-1}$ exists

To find adjoint of $A$

$A_{11}=0,A_{12}=2,A_{13}=1$

$A_{21}=-1,A_{22}=-9,A_{23}=-6$

$A_{31}=2,A_{32}=23,A_{33}=13$

$Adj\left(A\right)=\text{co-factor}\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -6\\ 2& 23& 13\end{array}\right]$

$=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -6& 13\end{array}\right]$

$A^{-1}=\frac{1}{|A|}Adj\left(A\right)$

$=\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -6& 13\end{array}\right]$

$X=A^{-1}B$

$=-\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -6& 13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

$X=-\left[\begin{array}{c}0+5-6\\ 22+45-69\\ 11+25-39\end{array}\right]$

$X=-\left[\begin{array}{c}-1\\ -2\\ -3\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

Hence, $x=1,y=2$ and $z=3$