If A is a square matrix of order 3, then, adj A, is equal to: Option 1: A Option 2: A, ^2 Option 3: A, ^3 Option 4: 3 A

Answer: A, ^2. Solution: For a square matrix of order n:, adj A, =, A, ^(n-1). For n = 3:, adj A, =, A, ^2.

Algebra

Matrices & Determinants

Easy

core

If A is a square matrix of order 3, then |adj A| is equal to:

$|A|$

$|A|^2$

$|A|^3$

$3|A|$

✅ Correct Option: 2

For a square matrix of order $n$ : $|\text{adj } A| = |A|^{n-1}$ . For $n = 3$ : $|\text{adj } A| = |A|^2$ .

2026: 19 May Shift 1

The matrix X such that it satisfies the equation: $X\begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ is

2026: 22 May Shift 2

If $f(x) = \begin{vmatrix} 0 & x+a & x+b \\ x-a & 0 & x+c \\ x-b & x-c & 0 \end{vmatrix}$ then the value of $f(0)$ is equal to:

2026: 22 May Shift 2

If $A^2 = 8A + kI, A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ then value of $k$ is:

2026: 30 May Shift 2

If A and B are symmetric matrices of same order, then which of the following statements are true?

(A) $AB^T+B^TA$ is a skew-symmetric matrix.

(B) $AB^T-BA^T$ is a skew-symmetric matrix

(D) $(AB)^{-1}(BA)^T=A$

Choose the correct answer from the options given below:

2026: 29 May Shift 2

Given that $[a_{ij}]$ is a square matrix of order n.

Match the LIST-I with LIST-II

LIST-I		LIST-II
A.	$A=[a_{ij}]; a_{ij}=\begin{cases}1, & \forall i=j\\0, & \forall i\neq j\end{cases}$	I. Identity Matrix
B.	$B=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i=j\\0, & \forall i\neq j\end{cases}$	II. Diagonal Matrix
C.	$C=[a_{ij}]; a_{ij}=\begin{cases}k, & \forall i=j\\0, & \forall i\neq j\end{cases}$	III. Scalar Matrix
D.	$D=[a_{ij}]; a_{ij}=\begin{cases}2i-3j, & \forall i\leq j\\0, & \forall i>j\end{cases}$	IV. Upper Triangular Matrix

Choose the correct answer from the options given below:

2026: 26 May Shift 2

The value/s of ' $\lambda$ ' for which the system of equations

$\lambda x + 3y - z = 1$

$x + 2y + z = 2$

$-\lambda x + y + 2z = -1$

has unique solution is/are:

(where $R$ is set of real numbers)