CUET · MATHS · PYQ PAPER 2025
Let \(P\) and \(Q\) be any two invertible matrices of the same order. Then Match List-I with List-II
| List - I (Matrix) | List - II (Equivalent matrix) |
| (A) \((P Q)^{-1}\) | (I) \(Q^{-1} P\) |
| (B) \(\left(P^{-1} Q\right)^{-1}\) | (II) \(Q P^{-1}\) |
| (C) \(\left(P Q^{-1}\right)^{-1}\) | (III) \(Q^{-1} P^{-1}\) |
| (D) \(\left(P^{-1} Q^{-1}\right)^{-1}\) | (IV) \(Q P\) |
- A (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
- B (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
- C (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
- D (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Answer & Solution
Correct Answer
(B) (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
Step-by-step Solution
Detailed explanation
(A) \((P Q)^{-1} = Q^{-1} P^{-1}\) (B) \((P^{-1} Q)^{-1} = Q^{-1} (P^{-1})^{-1} = Q^{-1} P\) (C) \((P Q^{-1})^{-1} = (Q^{-1})^{-1} P^{-1} = Q P^{-1}\) (D) \((P^{-1} Q^{-1})^{-1} = (Q^{-1})^{-1} (P^{-1})^{-1} = Q P\) Matches: (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
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