Algebra

Matrices & Determinants

Medium

applied

If $A=\begin{bmatrix}x+4 & 3\\2 & 1\end{bmatrix}, B=\begin{bmatrix}5 & 2x-1\\2 & x\end{bmatrix}$ and $\lvert adj(AB) \rvert=32$ , then the value of x is:

±3

±4

±6

±2

✅ Correct Option: 3

$|A|=(x+4)-6=x-2$ and $|B|=5x-2(2x-1)=x+2$ .

So $|AB|=x^2-4$ .

For order $2$ , $|adj(AB)|=|AB|^{2-1}=x^2-4=32$ , giving $x^2=36$ and $x=\pm6$ .

2025: 3 June Shift 1

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}$ , then

Match List-I with List-II

Choose the correct answer from the options given below:

2026: 30 May Shift 2

If A is square matrix of order 3 x 3 such that $a_{ij} = j - i$ , then which of the following statements are correct?

(A) A is a symmetric matrix.

(B) A is a skew symmetric matrix.

(D) A is a diagonal matrix.

Choose the correct answer from the options given below:

2026: 19 May Shift 2

If A and B are two square matrices of order 'n', where $AB \neq BA$ , then

2026: 22 May Shift 1

If $A=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 3\\1 & -2 & 1\end{bmatrix}$ , then

Match List-I with List-II

List-I	List-II
(A) $\lvert A\rvert$	(I) 48
(B) $\lvert 2A\rvert$	(II) 36
(C) $\left\lvert\operatorname{adj}(A)\right\rvert$	(III) 12
(D) $2\lvert A\rvert$	(IV) 6

Choose the correct answer from the options given below:

2022: 17 Aug Shift 2

Match List - I with List - II.

Choose the correct answer from the options given below :

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:

If A=[x+4321],B=[52x−12x]A=\begin{bmatrix}x+4 & 3\\2 & 1\end{bmatrix}, B=\begin{bmatrix}5 & 2x-1\\2 & x\end{bmatrix}A=[x+42​31​],B=[52​2x−1x​] and ∣adj(AB)∣=32\lvert adj(AB) \rvert=32∣adj(AB)∣=32, then the value of x is: