Let P and Q be any two invertible matrices of the same order. Then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Matrix | Equivalent matrix | ||
| (A) | \((P Q)^{-1}\) | (I) | \(Q^{-1}P\) |
| (B) | \((P^{-1} Q)^{-1}\) | (II) | \(QP^{-1}\) |
| (C) | \((P Q^{-1})^{-1}\) | (III) | \(Q^{-1} P^{-1}\) |
| (D) | \((P^{-1} Q^{-1})^{-1}\) | (IV) | \(QP\) |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Two results are used: the reversal law \((XY)^{-1} = Y^{-1}X^{-1}\), and \((X^{-1})^{-1} = X\).
(A) \((P Q)^{-1}\)
This matches (III) \(Q^{-1}P^{-1}\):
\((PQ)^{-1} = Q^{-1}P^{-1}\)
(B) \((P^{-1} Q)^{-1}\)
This matches (I) \(Q^{-1}P\):
\((P^{-1}Q)^{-1} = Q^{-1}(P^{-1})^{-1} = Q^{-1}P\)
(C) \((P Q^{-1})^{-1}\)
This matches (II) \(QP^{-1}\):
\((PQ^{-1})^{-1} = (Q^{-1})^{-1}P^{-1} = QP^{-1}\)
(D) \((P^{-1} Q^{-1})^{-1}\)
This matches (IV) \(QP\):
\((P^{-1}Q^{-1})^{-1} = (Q^{-1})^{-1}(P^{-1})^{-1} = QP\)
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).