Let $\mathrm{M}=(\mathrm{a}_{\mathrm{ij}}), \mathrm{i}, \mathrm{j} \in\{1,2,3\}$, be the $3 \times 3$ matrix such that $\mathrm{a}_{\mathrm{ij}}=1$ if $\mathrm{j}+1$ is divisible by i , otherwise $\mathrm{a}_{\mathrm{ij}}=0$. Then which of the following statements is (are) true ?

(B) There exists a nonzero column matrix $\left(\begin{array}{l}\mathrm{a}_{1} \\ \mathrm{a}_{2} \\ \mathrm{a}_{3}\end{array}\right)$ such that $\mathrm{M}\left(\begin{array}{l}\mathrm{a}_{1} \\ \mathrm{a}_{2} \\ \mathrm{a}_{3}\end{array}\right)=\left(\begin{array}{l}-\mathrm{a}_{1} \\ -\mathrm{a}_{2} \\ -\mathrm{a}_{3}\end{array}\right)$; (C) The set $\left\{\mathrm{X} \in \mathbb{R}^{3}: \mathrm{MX}=\mathbf{0}\right\} \neq\{\boldsymbol{0}\}$, where $\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)$

(D) The matrix (M - 2I) is invertible, where I is the $3 \times 3$ identity matrix

Matrices — JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 Question with Solution

Let

M = (a_{ij}), i, j \in {1, 2, 3}

, be the

3 \times 3

matrix such that

a_{ij} = 1

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 $M a_{1} a_{2} a_{3} = - a_{1} - a_{2} - a_{3}$ ✓ Correct
(C)The set ${X \in R^{3} : MX = 0} \neq = {0}$ , where $0 = 000$ ✓ Correct
(D)The matrix (M - 2I) is invertible, where I is the $3 \times 3$ identity matrix

Answer: (B), (C)

Solution

First construct matrix M. For

a_{ij} = 1

when

(j + 1)

is divisible by

i

M = 110101110

∣ M ∣ = 1 (0 - 1) - 1 (0 - 0) + 1 (1 - 0) = - 1 + 1 = 0

So M is singular (not invertible), option (A) is false.

For option (B):

M a_{1} a_{2} a_{3} = - a_{1} - a_{2} - a_{3}

gives eigenvalue

- 1

.
This gives:

a_{1} + a_{2} + a_{3} = - a_{1}

a_{1} + a_{3} = - a_{2}

a_{2} = - a_{3}

Solving:

a_{1} = 0

a_{2} = - a_{3}

. Nonzero solutions exist, so (B) is true.

For option (C):

M X = 0

gives

a_{1} + a_{2} + a_{3} = 0

a_{1} + a_{3} = 0

a_{2} = 0

.
This has infinitely many solutions (e.g.,

a_{1} = t, a_{2} = 0, a_{3} = - t

), so (C) is true.

For option (D):

M - 2 I = - 1 10 1 - 2 1 11 - 2

∣ M - 2 I ∣ = 0

, so not invertible. (D) is false.

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · One or more correct options

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Matrices

Solution

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