Let $A$ be a $3 \times 3$ real matrix such that

$$A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \text {. }$$

Then, the system $$(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$$ has :

<p>$$\text { Let } A=\left[\begin{array}{lll} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right]$$</p> <p>$$\text { Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right] \quad \text{ ..... (1)}$$</p> <p>$$\therefore\left[\begin{array}{l} \mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]$$</p> <p>$$\begin{aligned} \therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\ \mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\ \mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Given } A\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & \therefore\left[\begin{array}{l} -\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \end{aligned}$$</p> <p>$$\begin{array}{r} \Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\ -\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\ -\mathrm{x}_3+\mathrm{z}_3=4 \end{array}$$</p> <p>$$\begin{aligned} & \text { Given } A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \end{aligned}$$</p> <p>$$\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3 \end{aligned}$$</p> <p>$$\begin{gathered} \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ \therefore \text { Now }(A-31)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -1 \\ 2 \\ 3 \end{array}\right] \\ \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \\ {\left[\begin{array}{c} -z \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ {[z=-1],[y=-2],[x=-3]} \end{gathered}$$</p>