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

Let the matrix be A = \left[\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right]

Now A\left[$\begin{array}{ll}0 \\ 0 \\ 1\end{array}$\right] = \left[$\begin{array}{ll}{c}_{1} \\ {c}_{2} \\ {c}_{3}\end{array}$\right] = \left[$\begin{array}{ll}1 \\ 1 \\ 2\end{array}$\right]

$⇒ {c}_{1} = 1, {c}_{2} = 1, {c}_{3} = 2$

A\left[$\begin{array}{ll}1 \\ 0 \\ 1\end{array}$\right] = \left[$\begin{array}{ll}{c}_{1} + {a}_{1} \\ {c}_{2} + {a}_{2} \\ {c}_{3} + {a}_{3}\end{array}$\right] = \left[$\begin{array}{ll}- 1 \\ 0 \\ 1\end{array}$\right]

$⇒ {a}_{1} = - 2, {a}_{2} = - 1, {a}_{3} = - 1$

A\left[$\begin{array}{ll}1 \\ 1 \\ 0\end{array}$\right] = \left[$\begin{array}{ll}{a}_{1} + {b}_{1} \\ {a}_{2} + {b}_{2} \\ {a}_{3} + {b}_{3}\end{array}$\right] = \left[$\begin{array}{ll}1 \\ 1 \\ 0\end{array}$\right]

$⇒ {b}_{1} = 3, {b}_{2} = 2, {b}_{3} = 1$

So matrix A = \left[\begin{cases}- 2 & 3 & 1 \\ - 1 & 2 & 1 \\ - 1 & 1 & 2\end{cases}\right]

⇒ A - 2I = \left[\begin{cases}- 4 & 3 & 1 \\ - 1 & 0 & 1 \\ - 1 & 1 & 0\end{cases}\right]

i.e. \left|A - 2I\right| = 0$</p><p>Now$X = \left[$$\begin{array}{ll}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{array}$$\right]</p><p>So, \left[\begin{cases}- 4 & 3 & 1 \\ - 1 & 0 & 1 \\ - 1 & 1 & 0\end{cases}\right]\left[$\begin{array}{ll}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{array}$\right] = \left[$\begin{array}{ll}4 \\ 1 \\ 1\end{array}$\right]

$- 4{x}_{1} + 3{x}_{2} + {x}_{3} = 4 . . . \left(1\right)$

$- {x}_{1} + {x}_{3} = 1 . . . \left(2\right)$

$- {x}_{1} + {x}_{2} = 1 . . . \left(3\right)$

Solving the above system of equations we get infinite solutions.