Let M = (aij), i, j ∈ {1, 2, 3}, be the 3 × 3 matrix such that aij = 1 if j + 1 is divisible by i, otherwise aij = 0. Then which of the following statements is(are) true?

The correct true statements are:

1. There exists a nonzero column matrix [a₁; a₂; a₃] such that M[a₁; a₂; a₃] = [-a₁; -a₂; -a₃].

2.

The set {X ∈ ℝ³ : MX = 0} ≠ {0}, where 0=[0,0,0]

Step-by-step Explanation:

We are given a $3\times 3$ matrix $M=\left(a_{ij}\right)$ where $i,j\in \{1,2,3\}$ defined by:
$a_{ij}=1$ if $j+1$ is divisible by $i$, otherwise $a_{ij}=0$.

Let us evaluate each entry of $M$:

- Row 1 ($i=1$):
$j=1\Rightarrow 1+1=2$ (divisible by 1) $\Rightarrow a_{11}=1$
$j=2\Rightarrow 2+1=3$ (divisible by 1) $\Rightarrow a_{12}=1$
$j=3\Rightarrow 3+1=4$ (divisible by 1) $\Rightarrow a_{13}=1$

- Row 2 ($i=2$):
$j=1\Rightarrow 1+1=2$ (divisible by 2) $\Rightarrow a_{21}=1$
$j=2\Rightarrow 2+1=3$ (not divisible by 2) $\Rightarrow a_{22}=0$
$j=3\Rightarrow 3+1=4$ (divisible by 2) $\Rightarrow a_{23}=1$

- Row 3 ($i=3$):
$j=1\Rightarrow 1+1=2$ (not divisible by 3) $\Rightarrow a_{31}=0$
$j=2\Rightarrow 2+1=3$ (divisible by 3) $\Rightarrow a_{32}=1$
$j=3\Rightarrow 3+1=4$ (not divisible by 3) $\Rightarrow a_{33}=0$

Thus, matrix $M$ is:

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

1. Determinant of M:

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

Since $\left|M\right|=0$, matrix $M$ is singular (not invertible).
Also, since $\left|M\right|=0$, the system of equations $MX=0$ has non-trivial solutions. Therefore, the solution set $\{X\in {ℝ}^3:MX=0\}\ne \left\{0\right\}$ is TRUE.

2. Existence of a nonzero matrix X such that MX = -X:

The condition $MX=-X$ can be rewritten as $(M+I)X=0$.
Let us find matrix $M+I$:

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

Calculating its determinant:

$|M+I|=2(1-1)-1(1-0)+1(1-0)=0-1+1=0$

Since $|M+I|=0$, there exists a non-trivial solution vector $X={[a_1,a_2,a_3]}^T\ne 0$ satisfying $MX=-X$. Therefore, this statement is TRUE.