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Home Questions SU7FGVBRRB
Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

Let A be any invertible square matrix. Then

List-I

List-II

(A) $A-A^T$

(I) $|A|A^{-1}$

(B) $AA^T$

(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:

Options:

(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 Answer:

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

Explanation:

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Given: Let A be any invertible square matrix.

Matching List-I with List-II:

List-I List-II Explanation
(A) A − AT (II) Skew-symmetric $A - A^T$ is always skew-symmetric
(B) AAT (III) Symmetric $(AA^T)^T = A A^T$ ⇒ symmetric
(C) det(A−1) (IV) $[\det(A)]^{-1}$ $\det(A^{-1}) = \frac{1}{\det(A)}$
(D) adj A (I) $|A| A^{-1}$ $\text{adj}(A) = |A| \cdot A^{-1}$ for invertible A

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