CUET UG 2025 — Mathematics Algebra

2025mcqmedium

If $A = \begin{bmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{bmatrix}$ , then $|adj A|$ is equal to

Official previous-year question

Held on 14 May 2025 · Verified 13 Jul 2026.

Options

A
3
B
9
C
27
D
81

Did you get this right?

Solution

For any $n \times n$ matrix $A$ :

$|adj \, A| = |A|^{n-1}$

Since $A$ is a $3 \times 3$ matrix:

$|adj \, A| = |A|^{3-1}$

$= |A|^2$

Given: $A = \begin{bmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{bmatrix}$

Expanding along the first row:

$|A| = 0 \times C_{11} + 0 \times C_{12} + \sqrt{3} \times C_{13}$

Finding the cofactor $C_{13}$ by removing row 1 and column 3:

$C_{13} = (-1)^{1+3} \times \begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0 \end{vmatrix}$

$= (+1) \times (0 \times 0 - \sqrt{3} \times \sqrt{3})$

$= 1 \times (0 - 3)$

$= -3$

Therefore:

$|A| = \sqrt{3} \times (-3)$

$= -3\sqrt{3}$

$|adj \, A| = |A|^2$

$= (-3\sqrt{3})^2$

$= (-3)^2 \times (\sqrt{3})^2$

$= 9 \times 3$

$= 27$

Your note

CUET UG Algebra in other years

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Work through every CUET UG Algebra PYQ, year by year.

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CUET UG 2025 — Mathematics Algebra

2025mcqmedium

If $A = \begin{bmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{bmatrix}$ , then $|adj A|$ is equal to

Official previous-year question

Held on 14 May 2025 · Verified 13 Jul 2026.

Options

A
3
B
9
C
27
D
81

Did you get this right?

Solution

For any $n \times n$ matrix $A$ :

$|adj \, A| = |A|^{n-1}$

Since $A$ is a $3 \times 3$ matrix:

$|adj \, A| = |A|^{3-1}$

$= |A|^2$

Given: $A = \begin{bmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{bmatrix}$

Expanding along the first row:

$|A| = 0 \times C_{11} + 0 \times C_{12} + \sqrt{3} \times C_{13}$

Finding the cofactor $C_{13}$ by removing row 1 and column 3:

$C_{13} = (-1)^{1+3} \times \begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0 \end{vmatrix}$

$= (+1) \times (0 \times 0 - \sqrt{3} \times \sqrt{3})$

$= 1 \times (0 - 3)$

$= -3$

Therefore:

$|A| = \sqrt{3} \times (-3)$

$= -3\sqrt{3}$

$|adj \, A| = |A|^2$

$= (-3\sqrt{3})^2$

$= (-3)^2 \times (\sqrt{3})^2$

$= 9 \times 3$

$= 27$

Your note

CUET UG Algebra in other years

Algebra 2026 Algebra 2024 Algebra 2023 Algebra 2022

Keep practicing Algebra

Work through every CUET UG Algebra PYQ, year by year.

Browse all Mathematics topics

CUET UG 2025 — Mathematics Algebra

Options

Did you get this right?

Solution

Your note

CUET UG Algebra in other years

More CUET UG Algebra questions

Other Mathematics topics for CUET UG

Calculus

Applied-Mathematics

Statistics & Applications

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Algebra

CUET UG 2025 — Mathematics Algebra

Options

Did you get this right?

Solution

Your note

CUET UG Algebra in other years

More CUET UG Algebra questions

Other Mathematics topics for CUET UG

Calculus

Applied-Mathematics

Statistics & Applications

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Algebra