Question

Using matrix method, solve the following system of equation:
$x-2y=10,2x+y+3z=8$ and $-2y+z=7$

Answer 1

$x-2y=10.....\left(1\right)$

$2x+y+3z=8.....\left(2\right)$

$-2y+z=7.....\left(3\right)$

$A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right],X=\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right]$

$|A|=1(1+6)+2(2-0)+0$

$=7+4=11\ne 0$

$\therefore A^{-1}exists$

$a_{11}=\left[\begin{array}{cc}1& 3\\ -1& \end{array}\right]=(1+6)=7,a_{12}=-\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]=-2,a_{13}=\left[\begin{array}{cc}2& 1\\ 0& -2\end{array}\right]=-4$

$a_{21}=-\left[\begin{array}{cc}-2& 0\\ -2& 1\end{array}\right]=-(-2)=2,a_{22}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=1,a_{23}=-\left[\begin{array}{cc}1& -2\\ 0& -2\end{array}\right]=-(-2)=2$

$a_{31}=\left[\begin{array}{cc}-2& 0\\ 1& 3\end{array}\right]=-6,a_{32}=-\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]=-\left(3\right)=-3,a_{33}=\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]=1+4=5$

$\therefore adjA={\left[\begin{array}{ccc}7& -2& -4\\ 2& 1& 2\\ -6& -3& 5\end{array}\right]}^T$

$=\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]$

$A^{-1}=\frac{adjA}{|A|}$

$=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]$

$AX=B\Rightarrow X=A^{-1}B$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}70+16-42\\ -20+8-21\\ -40+16+35\end{array}\right]$

$=\frac{1}{11}\left[\begin{array}{c}44\\ -33\\ 11\end{array}\right]$

$\Rightarrow x=4,y=-3,z=1$