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:
Answer & explanation
Correct answer: option 4
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 |