Let A be a 3 × 3 real matrix such that A( array ll 1 0 1 array ) =...

Let, A = \left[\begin{cases}{x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ {x}_{3} & {y}_{3} & {z}_{3}\end{cases}\right]

It is given that, A\left[$\begin{array}{ll}1 \\ 0 \\ 1\end{array}$\right] = \left[$\begin{array}{ll}2 \\ 0 \\ 2\end{array}$\right]

⇒ \left[$\begin{array}{ll}{x}_{1} + {z}_{1} \\ {x}_{2} + {z}_{2} \\ {x}_{3} + {z}_{3}\end{array}$\right] = \left[$\begin{array}{ll}2 \\ 0 \\ 2\end{array}$\right]

$⇒ {x}_{1} + {z}_{1} = 2, {x}_{2} + {z}_{2} = 0, {x}_{3} + {z}_{3} = 0 . . . \left(i\right)$

Also, A\left[$\begin{array}{ll}- 1 \\ 0 \\ 1\end{array}$\right] = \left[$\begin{array}{ll}- 4 \\ 0 \\ 4\end{array}$\right]

⇒ \left[$\begin{array}{ll}- {x}_{1} + {z}_{1} \\ - {x}_{2} + {z}_{2} \\ - {x}_{3} + {z}_{3}\end{array}$\right] = \left[$\begin{array}{ll}- 4 \\ 0 \\ 4\end{array}$\right]

$⇒ - {x}_{1} + {z}_{1} = - 4, - {x}_{2} + {z}_{2} = 0, - {x}_{3} + {z}_{3} = 4 . . . \left(ii\right)$

Now, A\left[$\begin{array}{ll}0 \\ 1 \\ 0\end{array}$\right] = \left[$\begin{array}{ll}0 \\ 2 \\ 0\end{array}$\right]

⇒ \left[$\begin{array}{ll}{y}_{1} \\ {y}_{2} \\ {y}_{3}\end{array}$\right] = \left[$\begin{array}{ll}0 \\ 2 \\ 0\end{array}$\right]

$⇒ {y}_{1} = 0, {y}_{2} = 2, {y}_{3} = 0 . . . \left(iii\right)$

Using \left(i\right)\text{and}\left(ii\right)$,</p><p>$⇒ {x}_{1} = 3, {x}_{2} = 0, {x}_{3} = - 1$</p><p>$⇒ {z}_{1} = - 1, {z}_{2} = 0, {z}_{3} = 3$</p><p>$∴ A = \left[$\begin{cases}3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3\end{cases}$\right]</p><p>Now, \left(A - 3I\right)\left[$\begin{array}{ll}x \\ y \\ z\end{array}$\right] = \left[$\begin{array}{ll}- 1 \\ 2 \\ 3\end{array}$\right]

⇒ \left[\begin{cases}0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0\end{cases}\right]\left[$\begin{array}{ll}x \\ y \\ z\end{array}$\right] = \left[$\begin{array}{ll}- 1 \\ 2 \\ 3\end{array}$\right]

⇒ \left[$\begin{array}{ll}- z \\ - y \\ - x\end{array}$\right] = \left[$\begin{array}{ll}1 \\ 2 \\ 3\end{array}$\right]

$⇒ z = - 1, y = - 2, x = - 3$