Match List-I with List-II
|
List-I |
List-II |
|
(A) $\text{A(adj A)}$ |
(I) $\frac{1}{|A|}$ |
|
(B) $\text{|adj A|}$ |
(II) $|A|^n$ |
|
(C) $|A^{-1}|$ |
(III) $|A|I$ |
|
(D) $\text{|A(adj A)|}$ |
(IV) $|A|^{n-1}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Given:
(A) A(adj A)
Property: A(adj A) = |A| I → determinant is $|A(adj A)| = |A|^n$, but the expression itself equals |A|I. So, (A) matches (III)
(B) |adj A|
Property: Determinant of adjoint: $|adj A| = |A|^{n-1}$, where n is the order of matrix. So, (B) matches (IV)
(C) $|A^{-1}|$
Property: Determinant of inverse: $|A^{-1}| = 1/|A|$. So, (C) matches (I)
(D) |A(adj A)|
Property: $|A(adj A)| = |A|^n$. So, (D) matches (II)