Properties of Matrix Based on Divisibility Condition

To determine which of the given statements are true, we first construct the matrix $M = (a_{ij})_{3 \times 3}$, where $a_{ij} = 1$ if $j+1$ is divisible by $i$, and $a_{ij} = 0$ otherwise, for $i, j \in \{1, 2, 3\}$.

Evaluating the entries $a_{ij}$:

• For $i = 1$:

  • $j=1 \implies j+1 = 2$, divisible by $1 \implies a_{11} = 1$
  • $j=2 \implies j+1 = 3$, divisible by $1 \implies a_{12} = 1$
  • $j=3 \implies j+1 = 4$, divisible by $1 \implies a_{13} = 1$

• For $i = 2$:

  • $j=1 \implies j+1 = 2$, divisible by $2 \implies a_{21} = 1$
  • $j=2 \implies j+1 = 3$, not divisible by $2 \implies a_{22} = 0$
  • $j=3 \implies j+1 = 4$, divisible by $2 \implies a_{23} = 1$

• For $i = 3$:

  • $j=1 \implies j+1 = 2$, not divisible by $3 \implies a_{31} = 0$
  • $j=2 \implies j+1 = 3$, divisible by $3 \implies a_{32} = 1$
  • $j=3 \implies j+1 = 4$, not divisible by $3 \implies a_{33} = 0$

Thus, the matrix $M$ is: $M = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$

---

Step 1: Characteristic Polynomial of $M$

Let us find the characteristic equation $\det(M - \lambda I) = 0$: $\det(M - \lambda I) = \begin{vmatrix} 1-\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = 0$

Expanding along the third row: $\det(M - \lambda I) = -1 \cdot ((1-\lambda) \cdot 1 - 1 \cdot 1) + (-\lambda) \cdot ((1-\lambda)(-\lambda) - 1)$ $= -1(-\lambda) - \lambda(\lambda^2 - \lambda - 1)$ $= \lambda - \lambda^3 + \lambda^2 + \lambda$ $= -\lambda^3 + \lambda^2 + 2\lambda = -\lambda(\lambda^2 - \lambda - 2) = -\lambda(\lambda + 1)(\lambda - 2)$

Setting $\det(M - \lambda I) = 0$, the eigenvalues of $M$ are: $\lambda = 0, \quad \lambda = -1, \quad \lambda = 2$

---

Step 2: Verification of Options

• Option (A): Since $\lambda = 0$ is an eigenvalue, we have: $\det(M) = 0 \cdot (-1) \cdot 2 = 0$ Therefore, $M$ is not invertible. (Option A is FALSE)

• Option (B): The equation $MX = -X$ can be rewritten as $(M + I)X = \mathbf{0}$. This holds for any non-zero eigenvector corresponding to the eigenvalue $\lambda = -1$. Solving $(M + I)X = \mathbf{0}$: $\begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ Subtracting row 2 from row 1 gives $a_1 = 0$, and row 3 gives $a_2 + a_3 = 0 \implies a_3 = -a_2$. Choosing $a_2 = 1$, we get a non-zero column matrix $X = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$. (Option B is TRUE)

• Option (C): The set $\{X \in \mathbb{R}^3 : MX = \mathbf{0}\}$ is the null space of $M$. Since $\det(M) = 0$, there exist non-trivial (non-zero) solutions. For instance, setting $MX = \mathbf{0}$: $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0 \text{ and } x + z = 0$ A non-zero solution is $X = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$. Thus, $\{X \in \mathbb{R}^3 : MX = \mathbf{0}\} \neq \{\mathbf{0}\}$. (Option C is TRUE)

• Option (D): Since $\lambda = 2$ is an eigenvalue of $M$, we have: $\det(M - 2I) = 0$ Therefore, the matrix $(M - 2I)$ is singular and not invertible. (Option D is FALSE)

---

Conclusion

The correct statements are (B) and (C).