Let $$M = (a_{ij})$$, $$i, j \in \{1, 2, 3\}$$, 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?

The entry rule is: $$a_{ij}=1$$ if $$j+1$$ is divisible by $$i$$, otherwise $$a_{ij}=0$$, with $$i,j\in\{1,2,3\}$$.

Step 1: Constructing $$M$$
For each column index $$j$$ the number $$j+1$$ equals 2, 3, 4 respectively. Check divisibility by every row index $$i$$:

$$ \begin{array}{c|ccc} & j=1 & j=2 & j=3\\ j+1 & 2 & 3 & 4\\ \hline i=1 & 1 & 1 & 1\\ i=2 & 1 & 0 & 1\\ i=3 & 0 & 1 & 0 \end{array} $$

Hence
$$ M=\begin{pmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{pmatrix}. $$

Step 2: Determinant of $$M$$
Using the first row expansion,

$$ \det M =1\bigl|\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr| -1\bigl|\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\bigr| +1\bigl|\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\bigr| =1(-1)-1(0)+1(1)=0. $$

Since $$\det M=0$$, the rank of $$M$$ is <3 and $$M$$ is not invertible.

Step 3: Characteristic polynomial and eigenvalues

$$ \begin{aligned} \lvert M-\lambda I\rvert &=\begin{vmatrix} 1-\lambda&1&1\\ 1&-\lambda&1\\ 0&1&-\lambda \end{vmatrix}\\[2mm] &=(1-\lambda)(\lambda^2-1)-1(-\lambda)+1(1)\\ &=-\lambda^3+\lambda^2+2\lambda\\ &=-\lambda\bigl(\lambda^2-\lambda-2\bigr)\\ &=-\lambda(\lambda-2)(\lambda+1). \end{aligned} $$

Thus the eigenvalues are $$\lambda_1=0,\;\lambda_2=2,\;\lambda_3=-1$$.

Step 4: Verifying each option

Option A: $$\det M=0$$, so $$M$$ is not invertible ⇒ Option A is false.

Option B: The eigenvalue $$-1$$ exists, so there is a non-zero vector $$X$$ satisfying $$MX=-X$$. Hence Option B is true.

Option C: Because $$\det M=0$$, the null-space of $$M$$ is non-trivial, i.e. $$\{X\in\mathbb{R}^3:MX=\mathbf{0}\}\neq\{\mathbf{0}\}$$. Option C is true.

Option D: Since $$2$$ is an eigenvalue, $$\det(M-2I)=0$$, so $$M-2I$$ is not invertible. Option D is false.

Final result
Option B and Option C are correct.