CUET UG 2025 — Mathematics Algebra

2025mcqmedium

If matrix $A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix}$ and $|A^3| = 64$ , then the value of p is:

Official previous-year question

Held on 19 May 2025 · Verified 13 Jul 2026.

Options

A
$\pm 1$
B
$\pm 2$
C
$\pm 3$
D
$\pm 4$

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Solution

When a matrix is raised to a power, the determinant is also raised to that same power.

$|A^3| = |A|^3$

Given that $|A^3| = 64$ :

$|A|^3 = 64$

$|A| = \sqrt[3]{64}$

$|A| = 4$

For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $ad - bc$ .

For matrix $A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix}$ :

$|A| = (p)(p) - (-3)(-4)$

$|A| = p^2 - 12$

Since $|A| = 4$ :

$p^2 - 12 = 4$

$p^2 = 16$

$p = \pm 4$

Therefore, the value of $p$ is $\pm 4$ .

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Solution

When a matrix is raised to a power, the determinant is also raised to that same power.

$|A^3| = |A|^3$

Given that $|A^3| = 64$ :

$|A|^3 = 64$

$|A| = \sqrt[3]{64}$

$|A| = 4$

For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $ad - bc$ .

For matrix $A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix}$ :

$|A| = (p)(p) - (-3)(-4)$

$|A| = p^2 - 12$

Since $|A| = 4$ :

$p^2 - 12 = 4$

$p^2 = 16$

$p = \pm 4$

Therefore, the value of $p$ is $\pm 4$ .

CUET UG 2025 — Mathematics Algebra

Options

Did you get this right?

Solution

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CUET UG Algebra in other years

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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