JEE Advanced 2023 — Paper 2 — MATH-II solution

JEE Advanced2023Paper 2MATH-II

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?

A
$M$ is invertible
B
There exists a nonzero column matrix $a_{1} a_{2} a_{3}$ such that $M a_{1} a_{2} a_{3} = - a_{1} - a_{2} - a_{3}$
C
The set ${X \in R^{3} : M X = 0} \neq = {0}$ , where $0 = 000$
D
The matrix $(M - 2 I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix

Solution

From the divisibility rule, $M = 111101010 .$

Expanding along the third row, $det M = - 1 \cdot (1 \cdot 1 - 1 \cdot 1) + 0 = 0$ , so $M$ is singular. Option (A) is false and option (C) is true because $M X = 0$ has nontrivial solutions.

For (B), solve $M X = - X$ , i.e.\ $(M + I) X = 0$ . The system reduces to $a_{1} = 0$ and $a_{2} + a_{3} = 0$ , which has a one-parameter family of nonzero solutions, for example $(0, 1, - 1)^{T}$ . Option (B) is true.

For (D), $M - 2 I = - 111 1 - 21 01 - 2$ . Its determinant is $- 1 (4 - 1) - 1 (- 2 - 0) + 1 (1 - 0) = - 3 + 2 + 1 = 0$ , so $M - 2 I$ is not invertible. Option (D) is false.

$∴$ Correct options are (B) and (C).

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