Matrices & Determinants
Matrices and Determinants
Grade Class 12
Question:
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 ?<br>(A) $M$ is invertible<br>(B) There exists a nonzero column matrix $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ such that $M \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \end{pmatrix}$<br>(C) The set $\{X \in \mathbb{R}^3 : MX=0\} \neq \{0\}$, where $0 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$<br>(D) The matrix $(M - 2I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
(A) $M$ is invertible
(B) There exists a nonzero column matrix $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ such that $M \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \end{pmatrix}$
(C) The set $\{X \in \mathbb{R}^3 : MX=0\} \neq \{0\}$, where $0 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
(D) The matrix $(M - 2I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
Step-by-Step Solution
Key Concept: Construct the matrix M based on the given condition. Check for invertibility using the determinant. Check for eigenvalues to address options B and D. Check for null space to address option C.
The matrix $M$ is defined by $a_{ij} = 1$ if $i$ divides $j+1$.<br>For $i=1$: $1$ divides $j+1$ for $j=1, 2, 3$. So $a_{11}=1, a_{12}=1, a_{13}=1$.<br>For $i=2$: $2$ divides $j+1$. $j=1 \implies 2|2$ (True), $j=2 \implies 2|3$ (False), $j=3 \implies 2|4$ (True). So $a_{21}=1, a_{22}=0, a_{23}=1$.<br>For $i=3$: $3$ divides $j+1$. $j=1 \implies 3|2$ (False), $j=2 \implies 3|3$ (True), $j=3 \implies 3|4$ (False). So $a_{31}=0, a_{32}=1, a_{33}=0$.<br>$M = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$.<br>Determinant $|M| = 1(0-1) - 1(0-0) + 1(1-0) = -1 + 1 = 0$. Thus $M$ is not invertible. (A) is false.<br>Since $|M|=0$, there exists a non-zero $X$ such that $MX=0$. (C) is true.<br>For (B), we check if $-1$ is an eigenvalue of $M$. $|M+I| = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{vmatrix} = 2(1-1) - 1(1-0) + 1(1-0) = 0 - 1 + 1 = 0$. Since $|M - (-1)I| = 0$, $-1$ is an eigenvalue, so (B) is true.<br>For (D), $|M-2I| = \begin{vmatrix} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{vmatrix} = -1(4-1) - 1(-2-0) + 1(1-0) = -3 + 2 + 1 = 0$. Since the determinant is 0, $M-2I$ is not invertible. (D) is false.
Correct Answer: B,C