Let M = ( a _ ij ), i, j ∈ 1, 2, 3 , be the 3 × 3 matrix such that...

Given,

$M = \left({a}_{ij}\right), i, j \in { 1, 2, 3 }$

${a}_{ij} = 1\text{if}j + 1\text{is divisible by}i$

$\text{otherwise}{a}_{ij} = 0$

So, ${a}_{11} = 1, {a}_{12} = 1, {a}_{13} = 1, {a}_{21} = 1, {a}_{22} = 0 . . . . . . , {a}_{33} = 0$

So, M = \left[$\begin{array}{ll}111 \\ 101 \\ 010\end{array}$\right]

$⇒ | M | = 1 ( - 1 ) - 1 ( - 1 ) = - 1 + 1 = 0$

Hence, $M$ is not invertible

Now solving option \left(B\right)we get,</p><p>\left[\begin{cases}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{cases}\right]\left[$\begin{array}{ll}{a}_{1} \\ {a}_{2} \\ {a}_{3}\end{array}$\right] = \left[$\begin{array}{ll}- {a}_{1} \\ - {a}_{2} \\ - {a}_{3}\end{array}$\right]

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

Now on comparing we get,

${a}_{1} + {a}_{2} + {a}_{3} = - {a}_{1}, {a}_{1} + {a}_{3} = - {a}_{2} \text{ and } {a}_{2} = - {a}_{3}$

Now on solving we get, ${a}_{1} = 0 \text{ and } {a}_{2} + {a}_{3} = 0$

So, there will be infinite possibility for ${a}_{2} \text{ and } {a}_{3}$

$⇒$ There exist a column matrix (infinite possibilities)

Now solving option $\left(C\right)$ we get,
\left[$\begin{cases}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{cases}$\right]\left[$$\begin{array}{ll}x \\ y \\ z\end{array}$$\right] = \left[$$\begin{array}{ll}0 \\ 0 \\ 0\end{array}$$\right]$</p><p>$⇒ \left[$$\begin{array}{ll}x + y + z \\ x + z \\ y\end{array}$$\right] = \left[$$\begin{array}{ll}0 \\ 0 \\ 0\end{array}$$\right]$</p><p>Now on comparing we get,</p><p>$⇒ $$\begin{array}{ll}x + y + z = 0 \\ x + z = 0 \\ y = 0\end{array}$$$</p><p>Yes it is possible</p><p>Now solving option \left(D\right)$ we get,

| M - 2I | = \left[\begin{cases}- 1 & 1 & 1 \\ 1 & - 2 & 1 \\ 0 & 1 & - 2\end{cases}\right]

$⇒ | M - 2I | = - 1\left(3\right) - 1\left(- 2 - 1\right) = - 3 + 3 = 0$

Hence, it is not invertible.