If A = bmatrix & & 0 \\ - & & 0 \\ 0 & 0 & 1 bmatrix and B is a square matrix of order 3, then, AB, is equal to: Option 1: B, ^2 Option 2: B Option 3: ^2 B Option 4: ^2 B

Answer: B. Solution: Expanding along the third row/column:, A, = 1 * (^2 + ^2) = 1. Therefore, AB, =, A, , B, = 1 , B, =, B, .

Algebra

Matrices & Determinants

Medium

core

If $A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and B is a square matrix of order 3, then |AB| is equal to:

$|B|^2$

$|B|$

$\sin^2\theta |B|$

$\cos^2\theta |B|$

✅ Correct Option: 2

Expanding along the third row/column: $|A| = 1 \cdot (\cos^2\theta + \sin^2\theta) = 1$ . Therefore $|AB| = |A| \cdot |B| = 1 \cdot |B| = |B|$ .

2026: 19 May Shift 2

If $f(x) = \begin{vmatrix} 1 & x & x^2 \\ x & x^2 & 1 \\ x^2 & 1 & x \end{vmatrix}$ , then the value of $f(\sqrt[3]{3})$ is

2025: 26 May Shift 2

If a matrix $A = \begin{bmatrix} 5 & -8 \\ -3 & 5 \end{bmatrix}$ then which of the following is / are TRUE?

(A) $|A| = 1$

(B) $A$ is a singular matrix.

(D) $AI = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$ $I$ is an identity matrix of order 2.

Choose the correct answer from the options given below:

2025: 30 May Shift 2

Given a matrix A of order 3x3. If |A|=3 then the value of |A(adj A)| is:

2025: 15 May Shift 1

Let A be a square matrix of order 2 such that $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , then A is:

2025: 2 June Shift 1

If $A$ is an invertible matrix of order 3, then

Match List-I with List-II

List-I	List-II
(A) $\vert \text{adj} A\vert$	(I) $8\vert A\vert$
(B) $\vert A(\text{adj} A)\vert$	(II) $\vert A\vert ^2$
(C) $\vert 2A\vert$	(III) $\frac{1}{\vert A\vert }$
(D) $\vert A^{-1}\vert$	(IV) $\vert A\vert ^3$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

If $A=\begin{bmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{bmatrix}$ , $\lambda\ne0$ , then $|adjA|$ is equal to

If A=[cos⁡θsin⁡θ0−sin⁡θcos⁡θ0001]A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}A=​cosθ−sinθ0​sinθcosθ0​001​​ and B is a square matrix of order 3, then |AB| is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and B is a square matrix of order 3, then |AB| is equal to: