If matrix A = bmatrix p & -3 \\ -4 & p bmatrix and, A^3, = 64, then the value of p is: Option 1: ± 1 Option 2: ± 2 Option 3: ± 3 Option 4: ± 4

Answer: ± 4. 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, = 3√64, A, = 4 For a 2×2 matrix bmatrix a & b \\ c & d bmatrix, the determinant is ad - bc. For matrix A = bmatrix p & -3 \\ -4 & p bmatrix:, A, = (p)(p) - (-3)(-4), A, = p^2 - 12 Since, A, = 4: p^2 - 12 = 4 p^2 = 16 p = ± 4 Therefore, the value of p is ± 4.

Algebra

Matrices & Determinants

Medium

core

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

$\pm 1$

$\pm 2$

$\pm 3$

$\pm 4$

✅ Correct Option: 4

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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List-I	List-II
(Matrix A)	(Determinant of adj A)
(A) $\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix}$	(I) 6
(B) $\begin{bmatrix} 0 & 1 \\ 4 & -1 \end{bmatrix}$	(II) 5
(C) $\begin{bmatrix} 1 & 2 \\ -3 & -1 \end{bmatrix}$	(III) -4
(D) $\begin{bmatrix} 4 & -2 \\ 3 & 0 \end{bmatrix}$	(IV) -2

Choose the correct answer from the options given below:

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If area of the triangle formed by A(1, 3), B(0, 0) and C(k, 0) is 3 sq. units, then which of the following is/are TRUE?

A. k = 2

B. k = -2

C. k = 3

D. k = -3

Choose the correct answer from the options given below:

If matrix A=[p−3−4p]A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix}A=[p−4​−3p​] and ∣A3∣=64|A^3| = 64∣A3∣=64, then the value of p is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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