Let $M=\left(a_{ij}\right), i, j\in \{1,2,3\}$, be the $3\times 3$ matrix such that $a_{ij}=1$ if $j+1$ is divisible by $i$, otherwise $a_{ij}=0$. Then which of the following statements is(are) true?

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, $M=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right]$

$\Rightarrow \left|M\right|=1(-1)-1(-1)=-1+1=0$

Hence, $M$ is not invertible

Now solving option $\left(B\right)$ we get,

$\left[\begin{array}{lll}1& 1& 1\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{l}{a}_1\\ a_2\\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1\\ -a_2\\ -a_3\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}{a}_1+a_2+a_3\\ a_1+a_3\\ a_2\end{array}\right]=\left[\begin{array}{c}-a_1\\ -a_2\\ -a_3\end{array}\right]$

Now on comparing we get,

$a_1+a_2+a_3=-a_1, a_1+a_3=-a_2 \& a_2=-a_3$

Now on solving we get, $a_1=0 \& a_2+a_3=0$

So, there will be infinite possibility for $a_2 \& a_3$

$\Rightarrow$ There exist a column matrix (infinite possibilities)

Now solving option $\left(C\right)$ we get,
$\left[\begin{array}{lll}1& 1& 1\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}x+y+z\\ x+z\\ y\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Now on comparing we get,

$\Rightarrow \begin{array}{c}x+y+z=0\\ x+z=0\\ y=0\end{array}$

Yes it is possible

Now solving option $\left(D\right)$ we get,

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

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

Hence, it is not invertible.