Given $A\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]$ ....$(1)$

$\therefore\left[\begin{array}{c}\mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]$

$\therefore \mathrm{x}_1+\mathrm{z}_1  =2$ $..........(2)$
$\mathrm{x}_2+\mathrm{z}_2  =0 $  $..............(3)$
$\mathrm{x}_3+\mathrm{z}_3 =0$ $..............(4)$

Given $A\left[\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}-4 \\ 0 \\ 4\end{array}\right]$

$\therefore\left[\begin{array}{l}-\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 4\end{array}\right]$

$\Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 $              $........(5)$

$-\mathrm{x}_2+\mathrm{x}_2=0 $                      $........(6)$

$-\mathrm{x}_3+\mathrm{z}_3=4$

Given $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$

$\therefore\left[\begin{array}{l}\mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$

$\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text { from (2), (3), (4), (5), (6) and (7) } \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3\end{aligned}$

$\begin{gathered}\therefore A=\left[\begin{array}{ccc}3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3\end{array}\right] \\ \therefore \operatorname{Now}(A-31)\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}-1 \\ 2 \\ 3\end{array}\right]\end{gathered}$

$\begin{aligned} & \therefore\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right] \\ & {\left[\begin{array}{l}-z \\ -y \\ -x\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]} \\ & {[z=-1],[y=-2],[x=-3]}\end{aligned}$