Match List-I with List-II Let A be any invertible square matrix. Then, List-I, List-II, (A) A - A^T, (I) A A^(-1), (B) A A^T, (II) Skew-symmetric, (C) (A^(-1)), (III) Symmetric, (D) adj A, (IV) [(A)]^(-1), Choose the correct answer from the options given below: Option 1: (A) - (III), (B) - (II), (C) - (IV), (D) - (I) Option 2: (A) - (III), (B) - (II), (C) - (I), (D) - (IV) Option 3: (A) - (II), (B) - (III), (C) - (I), (D) - (IV) Option 4: (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

Answer: (A) - (II), (B) - (III), (C) - (IV), (D) - (I). Solution: (A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T) Since the transpose equals the negative of itself, A - A^T is skew-symmetric. (A) arrow (II) (AA^T)^T = (A^T)^T * A^T = A * A^T = AA^T Since the transpose equals itself, AA^T is symmetric. (B) arrow (III) Since A * A^(-1) = I, taking determinant on both sides: (A) * (A^(-1)) = (I) = 1 (A^(-1)) = 1/((A)) = [(A)]^(-1) (C) arrow (IV) From the standard identity: A^(-1) = (adj(A))/((A)) adj(A) = (A) * A^(-1) =, A, A^(-1) (D) arrow (I), List-I, List-II, (A) A - A^T, (II) Skew-symmetric, (B) AA^T, (III) Symmetric, (C) (A^(-1)), (IV) [(A)]^(-1), (D) adj(A), (I) \, A\, A^(-1)

Algebra

Matrices & Determinants

Easy

core

Match List-I with List-II

Let $A$ be any invertible square matrix. Then

List-I List-II
(A) $A - A^T$ (I) $\vert A\vert A^{-1}$
(B) $A A^T$ (II) Skew-symmetric
(C) $\det (A^{-1})$ (III) Symmetric
(D) $\text{adj} A$ (IV) $[\det(A)]^{-1}$

Choose the correct answer from the options given below:

List-I	List-II
(A) $A - A^T$	(I) $\vert A\vert A^{-1}$
(B) $A A^T$	(II) Skew-symmetric
(C) $\det (A^{-1})$	(III) Symmetric
(D) $\text{adj} A$	(IV) $[\det(A)]^{-1}$

(A) - (III), (B) - (II), (C) - (IV), (D) - (I)

(A) - (III), (B) - (II), (C) - (I), (D) - (IV)

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

✅ Correct Option: 4

$(A - A^T)^T = A^T - (A^T)^T$

$= A^T - A$

$= -(A - A^T)$

Since the transpose equals the negative of itself, $A - A^T$ is skew-symmetric.

$(A) \rightarrow \text{(II)}$

$(AA^T)^T = (A^T)^T \cdot A^T$

$= A \cdot A^T$

$= AA^T$

Since the transpose equals itself, $AA^T$ is symmetric.

$(B) \rightarrow \text{(III)}$

Since $A \cdot A^{-1} = I$ , taking determinant on both sides:

$\det(A) \cdot \det(A^{-1}) = \det(I) = 1$

$\det(A^{-1}) = \dfrac{1}{\det(A)}$

$= [\det(A)]^{-1}$

$(C) \rightarrow \text{(IV)}$

From the standard identity:

$A^{-1} = \dfrac{\text{adj}(A)}{\det(A)}$

$\text{adj}(A) = \det(A) \cdot A^{-1}$

$= |A|\,A^{-1}$

$(D) \rightarrow \text{(I)}$

List-I	List-II
(A) $A - A^T$	(II) Skew-symmetric
(B) $AA^T$	(III) Symmetric
(C) $\det(A^{-1})$	(IV) $[\det(A)]^{-1}$
(D) $\text{adj}(A)$	(I) $\\|A\\|A^{-1}$

2025: 16 May Shift 1

If A is a square matrix of order 3 and |A| = -5, then |3A| is equal to

2025: 30 May Shift 2

If $A = [a_{ij}]_{3 \times 2}$ where $a_{ij} = i + j$ , then

(A) A is a square matrix

(B) $a_{21} + a_{32} = 8$

(D) Transpose of $A = \begin{bmatrix}2 & 3 \\ 3 & 4 \\ 4 & 5\end{bmatrix}$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

Let A be the area of the triangle with vertices (3, 8), (5, -1) and $(k, 2)$ , $k \in \mathbb{Z}$ (set of integers). If $2A = 75$ , then k is equal to:

2023: 15 June Shift 2

If A is a square matrix of order 3 and $|A| = 5$ , then $|adj(adjA)|$ is :

2025: 30 May Shift 2

The diagonal elements of a skew symmetric matrix are all

2023: 15 June Shift 2

The value of $2y - 3x$ , if $2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$ is :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)