​Let A be a 3 x 3 real matrix such that

Question

Let \(A\) be a \(3 \times 3\) real matrix such that

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

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

(1) unique solution

(2) exactly two solutions

(3) no solution

(4) infinitely many solutions

Answer 1

Correct option is (1) unique solution

Let \(A=\left[\begin{array}{lll}x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3}\end{array}\right]\)

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

\(\therefore\left[\begin{array}{l}\mathrm{x}_{1}+\mathrm{z}_{1} \\ \mathrm{x}_{2}+\mathrm{z}_{2} \\ \mathrm{x}_{3}+\mathrm{z}_{3}\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]\)

\(\therefore \mathrm{x}_{1}+\mathrm{z}_{1}=2 \quad....(2)\)

\(\mathrm{x}_{2}+\mathrm{z}_{2}=0 \quad....(3)\)

\(\mathrm{x}_{3}+\mathrm{z}_{3}=0 \quad....(4)\)

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

\(\therefore\left[\begin{array}{c}-\mathrm{x}_{1}+\mathrm{z}_{1} \\ -\mathrm{x}_{2}+\mathrm{z}_{2} \\ -\mathrm{x}_{3}+\mathrm{z}_{3}\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 4\end{array}\right]\)

\(\Rightarrow-\mathrm{x}_{1}+\mathrm{z}_{1}=-4 \quad....(5)\)

\(-\mathrm{x}_{2}+\mathrm{x}_{2}=0 \quad ....(6)\)

\(-\mathrm{x}_{3}+\mathrm{z}_{3}=4\)

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

\(\therefore\left[\begin{array}{l}\mathrm y_1 \\\mathrm y_2 \\ \mathrm y_3\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right] \)

\(\therefore \mathrm{y}_{1}=0, \mathrm{y}_{2}=2, \mathrm{y}_{3}=0\)

\(\therefore\) from (2), (3), (4), (5), (6) and (7)

\(\mathrm{x}_{1}=3 \mathrm{x}, \mathrm{x}_{2}=0, \mathrm{x}_{3}=-1\)

\(\mathrm{y}_{1}=0, \mathrm{y}_{2}=2, \mathrm{y}_{3}=0\)

\(\mathrm{z}_{1}=-1, \mathrm{z}_{2}=0, \mathrm{z}_{3}=3\)

\(\therefore A=\left[\begin{array}{ccc}3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3\end{array}\right]\)

\(\therefore \text{ Now } (A-31)\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}-1 \\ 2 \\ 3\end{array}\right]\)

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

\(\left[\begin{array}{l}-\mathrm{z} \\ -\mathrm{y} \\ -\mathrm{x}\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\)

\([\mathrm{z}=-1],[\mathrm{y}=-2],[\mathrm{x}=-3]\)