If A^T = bmatrix -2 & 3 \\ 1 & 2 bmatrix and B = bmatrix -1 & 0 \\ 1 & 2 bmatrix, then the matrix (A + 2B)^T is Option 1: bmatrix 4 & 5 \\ -1 & -6 bmatrix Option 2: bmatrix -4 & -5 \\ 1 & 6 bmatrix Option 3: bmatrix -4 & 5 \\ 1 & 6 bmatrix Option 4: bmatrix 4 & -5 \\ -1 & 6 bmatrix

Answer: bmatrix -4 & 5 \\ 1 & 6 bmatrix. Solution: Given A^T = bmatrix -2 & 3 \\ 1 & 2 bmatrix and B = bmatrix -1 & 0 \\ 1 & 2 bmatrix. To find (A + 2B)^T, matrix A is needed first. Since A^T is given, transpose it to find A: A = (A^T)^T A = bmatrix -2 & 3 \\ 1 & 2 bmatrix^T A = bmatrix -2 & 1 \\ 3 & 2 bmatrix Calculate 2B: 2B = 2 x bmatrix -1 & 0 \\ 1 & 2 bmatrix 2B = bmatrix -2 & 0 \\ 2 & 4 bmatrix Calculate A + 2B: A + 2B = bmatrix -2 & 1 \\ 3 & 2 bmatrix + bmatrix -2 & 0 \\ 2 & 4 bmatrix A + 2B = bmatrix -4 & 1 \\ 5 & 6 bmatrix Calculate (A + 2B)^T: (A + 2B)^T = bmatrix -4 & 1 \\ 5 & 6 bmatrix^T (A + 2B)^T = bmatrix -4 & 5 \\ 1 & 6 bmatrix Therefore, (A + 2B)^T = bmatrix -4 & 5 \\ 1 & 6 bmatrix

If A^T = bmatrix -2 & 3 \\ 1 & 2 bmatrix and B = bmatrix -1 & 0 \\ 1 & 2 bmatrix, then the matrix (A + 2B)^T is

bmatrix -4 & 5 \\ 1 & 6 bmatrix

bmatrix 4 & 5 \\ -1 & -6 bmatrix

bmatrix -4 & -5 \\ 1 & 6 bmatrix

bmatrix -4 & 5 \\ 1 & 6 bmatrix

bmatrix 4 & -5 \\ -1 & 6 bmatrix

Algebra

Matrices & Determinants

Easy

common

If $A^T = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$ , then the matrix $(A + 2B)^T$ is

$\begin{bmatrix} 4 & 5 \\ -1 & -6 \end{bmatrix}$

$\begin{bmatrix} -4 & -5 \\ 1 & 6 \end{bmatrix}$

$\begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

$\begin{bmatrix} 4 & -5 \\ -1 & 6 \end{bmatrix}$

✅ Correct Option: 3

Given $A^T = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$ .

To find $(A + 2B)^T$ , matrix $A$ is needed first.

Since $A^T$ is given, transpose it to find $A$ :

$A = (A^T)^T$

$A = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}^T$

$A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$

Calculate $2B$ :

$2B = 2 \times \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$

$2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

Calculate $A + 2B$ :

$A + 2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

$A + 2B = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$

Calculate $(A + 2B)^T$ :

$(A + 2B)^T = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}^T$

$(A + 2B)^T = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

Therefore, $(A + 2B)^T = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

2026: 22 May Shift 2

Which of the following statements are true?

(A) A matrix which is both symmetric as well as skew-symmetric is a null matrix.

(B) All positive integral powers of a symmetric matrix are symmetric.

(D) All positive even integral powers of a skew-symmetric matrix are skew-symmetric.

Choose the correct answer from the options given below:

2026: 26 May Shift 2

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	Column matrix	I.	$\begin{bmatrix}1 & 2 & 3\end{bmatrix}$
B.	Row matrix	II.	$\begin{bmatrix}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{bmatrix}$
C.	Diagonal matrix	III.	$\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
D.	Scalar matrix	IV.	$\begin{bmatrix}2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{bmatrix}$

Choose the correct answer from the options given below:

2026: 25 May Shift 1

For a matrix $A = \begin{bmatrix} a & -1 & b \\ -2 & 1 & -1 \\ 1 & 0 & 2 \end{bmatrix}$ , $C_{32} = -1$ and $C_{33} = 3$ , where $C_{ij}$ is the cofactor of element $a_{ij}$ in $A$ . Then

2025: 13 May Shift 1

Match List-I with List-II

List-I	List-II
Type of matrix	Conditions
(A) Square matrix A	(I) $A = [a_{ij}]_{m \times m}$ where $\begin{cases} a_{ij} = 0 & , i \neq j \\ a_{ij} = k & , i = j \end{cases}$ , where $k \neq 0$ is constant.
(B) Scalar Matrix A	(II) $A = [a_{ij}]_{m \times m}$
(C) Diagonal matrix A	(III) $A = [a_{ij}]_{m \times m}$ where $\begin{cases} a_{ij} = 0 & , i \neq j \\ a_{ij} = 1 & , i = j \end{cases}$
(D) Identity matrix A	(IV) $A = [a_{ij}]_{m \times m}$ where $a_{ij} = 0$ , $i \neq j$

Choose the correct answer from the options given below:

2023: 22 May Shift 3

The value of the determinant $\Delta = \begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix}$ is :

2026: 29 May Shift 2

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $\begin{vmatrix} x+2 & 3 \\ x+5 & 4 \end{vmatrix} = 3$	I. $x = 10$
B. $\begin{vmatrix} x+2 & x+5 \\ 3 & 4 \end{vmatrix} = 13$	II. $x = 60$
C. $\begin{vmatrix} x+4 & x-4 \\ 8 & 7 \end{vmatrix} = 0$	III. $x = 0$
D. $\begin{vmatrix} x+4 & 8 \\ x-4 & 7 \end{vmatrix} = 60$	IV. $x = 20$

Choose the correct answer from the options given below:

If AT=[−2312]A^T = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}AT=[−21​32​] and B=[−1012]B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}B=[−11​02​], then the matrix (A+2B)T(A + 2B)^T(A+2B)T is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $A^T = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$ , then the matrix $(A + 2B)^T$ is