JEE Advanced 2023 · Paper 2 · Mathematics Q5

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MathematicsMultiple correct+4 / -2Question 5 of 51

Let $M = (a_{ij})$ , $i, j \in \{1,2,3\}$ , be the $3\times3$ 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\times3

identity matrix

Answer and worked solution

Official answer: B, C

The matrix. $a_{ij}=1$ when $i$ divides $j+1$ :

row 1: all 1s;
row 2: $j=1,3$ ;
row 3: $j=2$ .
$M=\begin{pmatrix}1&1&1\\1&0&1\\0&1&0\end{pmatrix},\qquad\det M=1(0-1)-1(0-0)+1(1-0)=0.$

(A) False and (C) True. $M$ is singular, so $MX=\mathbf0$ has non-zero solutions, e.g. $X=(1,0,-1)^T$ .

(B) True.
$\det(M+I)=\begin{vmatrix}2&1&1\\1&1&1\\0&1&1\end{vmatrix}=0,$
so $MX=-X$ has a non-zero solution, e.g. $X=(0,1,-1)^T$ .

(D) False.
$\det(M-2I)=\begin{vmatrix}-1&1&1\\1&-2&1\\0&1&-2\end{vmatrix}=-3+2+1=0,$
so $M-2I$ is singular.

Topics:Determinants and properties Systems of linear equations

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JEE Advanced 2023 · Paper 2, Mathematics question 5