Sol. $$ \begin{aligned} & \text { Let } A=\left[\begin{array}{lll} x_1 & y_1 & x_1 \\ x_2 & y_2 & x_2 \\ x_3 & y_2 & x_3 \end{array}\right] \text { Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]=\left[\begin{array}{l} x_4+x_1 \\ x_4+x_2 \\ x_4+x_2 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]=x_1+x_1=2 \\ & B_8+B_4=0 \\ & 5_1+4_2=0 \\ & \text { Grean A }\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & 4\left[\begin{array}{l} -x_3+x_4 \\ -x_3+x_3 \\ -x_3+x_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \\ & =-B_1+B_1=-4 \\ & -x_2+x_2=0 \\ & -x_3+x_4=4 \\ & \therefore y_1=2, y_2=2, y_2=0 \\ & \text { A from (2), (3), (4), [5], (6) and (7) } \end{aligned} $$ $$ \begin{aligned} & y_1=0, y_2=2, y_2=0 \\ & z_1=-1, z_2=0, z_2=3 \\ & \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ & \therefore \operatorname{Now}(A-31)\left[\begin{array}{l} x_n \\ l \end{array}\right]=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right] \\ & \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x_n \\ l \end{array}\right]=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right] \\ & {\left[\begin{array}{l} -\frac{1}{2} \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ & \mid x=-1\}|y=-2|,|x=-3| \end{aligned} $$