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?
$M$ is invertible
There exists a nonzero column matrix $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ such that $M \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \end{pmatrix}$
The set $\{X \in \mathbb{R}^3 : MX = \mathbf{0}\} \neq \{\mathbf{0}\}$, where $\mathbf{0} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
The matrix $(M - 2I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
Step-by-Step Solution
Key Concept: Applying properties of dot and cross products to find vector magnitude and angles, and using scalar triple products.
**Step 1: Construct the matrix M**
Let's determine the elements $a_{ij}$ for $1 \le i, j \le 3$. $a_{ij} = 1$ if $i$ divides $j+1$, else $0$.\nRow 1 ($i=1$): $j+1 \in \{2, 3, 4\}$. 1 divides all of them, so $a_{11}=1, a_{12}=1, a_{13}=1$.\nRow 2 ($i=2$): $j+1 \in \{2, 3, 4\}$. 2 divides 2 and 4, so $j=1$ and $j=3$. Thus $a_{21}=1, a_{22}=0, a_{23}=1$.\nRow 3 ($i=3$): $j+1 \in \{2, 3, 4\}$. 3 divides 3, so $j=2$. Thus $a_{31}=0, a_{32}=1, a_{33}=0$.\nSo, $M = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$.
**Step 2: Evaluate invertibility of M**
Calculate the determinant of $M$: $|M| = 1(0 - 1) - 1(0 - 0) + 1(1 - 0) = -1 + 1 = 0$. Since $|M| = 0$, $M$ is not invertible. This makes (A) false. Because $|M|=0$, the equation $MX = \mathbf{0}$ has non-trivial solutions, which makes (C) true.
**Step 3: Find eigenvalues of M**
The characteristic equation is $|M - \lambda I| = 0$.\n$\begin{vmatrix} 1-\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = (1-\lambda)(\lambda^2 - 1) - 1(-\lambda) + 1(1) = 0$.\n$(1-\lambda)(\lambda-1)(\lambda+1) + \lambda + 1 = 0 \implies -(\lambda-1)^2(\lambda+1) + (\lambda+1) = 0$.\n$(\lambda+1)[-(\lambda^2 - 2\lambda + 1) + 1] = 0 \implies (\lambda+1)(-\lambda^2 + 2\lambda) = 0 \implies -\lambda(\lambda+1)(\lambda-2) = 0$.\nSo eigenvalues are $\lambda = 0, -1, 2$.
**Step 4: Evaluate remaining options**
For option (B), the equation $MX = -X$ means $-1$ is an eigenvalue of $M$. Since $-1$ is an eigenvalue, there exists a non-zero eigenvector, so (B) is true.\nFor option (D), the eigenvalues of $M - 2I$ are $\lambda - 2$, which are $-2, -3, 0$. Since $0$ is an eigenvalue, $|M-2I|=0$, meaning it is not invertible, so (D) is false.
Correct Answer: 2, 3