JEE Advanced 2023 Paper 2 Q05 Mathematics Matrices & Determinants Matrix Operations Medium

JEE Advanced 2023 Paper 2 · Q05 · Matrix Operations

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?

  1. A. $M$ is invertible
  2. B. There exists a nonzero column matrix $\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ such that $M\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} = \begin{pmatrix}-a_1\\-a_2\\-a_3\end{pmatrix}$
  3. C. The set $\{X \in \mathbb{R}^3 : MX = \vec{0}\} \ne \{\vec{0}\}$
  4. D. The matrix $(M - 2I)$ is invertible, where $I$ is the $3\times 3$ identity matrix
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C

Solution

Construct $M$: row $i=1$ (any $j+1$ divisible by 1) → all 1s; row $i=2$ ($j=1,3$ since $2,4$ divisible by 2) → $(1,0,1)$; row $i=3$ ($j=2$ since $3$ divisible by 3) → $(0,1,0)$. So $M = \begin{pmatrix}1&1&1\\1&0&1\\0&1&0\end{pmatrix}$. $\det M = 1(0-1) - 1(0-0) + 1(1-0) = -1+1 = 0$, so $M$ is singular — (A) FALSE, (D) check $|M-2I|$: $\det\begin{pmatrix}-1&1&1\\1&-2&1\\0&1&-2\end{pmatrix} = -1(4-1) - 1(-2-0) + 1(1-0) = -3+2+1 = 0$ — also singular, so (D) FALSE. (B) Solving $(M+I)X = 0$: yields infinite solutions (the system is consistent and rank-deficient) — TRUE. (C) Since $\det M = 0$, kernel is non-trivial — TRUE.

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