Solve that equation

Solution:

Given, $\quad A=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$

Co-factors are:

$A_{11}=-3, A_{12}=2, A_{13}=2$

$A_{21}=-2, A_{22}=1, A_{23}=1$

$A_{31}=-4, A_{32}=2, A_{33}=3$

Now,

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

$|A|=1(-3)-2(-2)+0=1$

Hence,

$A^{-1}=\frac{\operatorname{adj} A}{|A|}=\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]$

Now, the system of linear equations are

$x-2 y=10$

$2 x-y-z=8$

and, $\quad-2 y+z=7$

Or $A X=B$

$\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

where, $A=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $B+\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

Thus, $X=A^{-1} B$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]=\left[\begin{array}{c}-30+16+14 \\ -20+8+7 \\ -40+16+21\end{array}\right]=\left[\begin{array}{c}0 \\ -5 \\ -3\end{array}\right]$

$\therefore x=0, y=-5$ and $z=-3$