CUET · MATHS · PYQ PAPER 2025
Match List-I with List-II
| List-I | List-II |
| (A) If A is a non singular matrix of order n, then \(| A (\operatorname{adjA})|\) is equal to | (I) \(|A|^{n-1}\) |
| (B) If A is a non singular matrix of order n, then |adj(adj A)| is equal to | (II) \(|A|^{n-2} A\) |
| (C) If A is a non singular matrix of order n, then \(\operatorname{adj}(\operatorname{adj} A )\) is equal to | (III) \(|A|^n\) |
| (D) if A is a non singular matrix of order n, then \(|(\operatorname{adj} A)|\) is equal to | (IV) \(|A|^{(n-1)^2}\) |
- A (A) - (I), (B) - (II), (С) - (III), (D) – (IV)
- B (А) - (II), (B) – (I), (C) - (III), (D) - (IV)
- C (А) - (III), (В) - (IV), (C) - (II), (D) - (I)
- D (А) - (III), (В) - (IV), (C) - (I), (D) - (II)
Answer & Solution
Correct Answer
(C) (А) - (III), (В) - (IV), (C) - (II), (D) - (I)
Step-by-step Solution
Detailed explanation
(A) \(| A (\operatorname{adjA})| = ||A|I| = |A|^n |I| = |A|^n\) Match: (III) (B) \(|\operatorname{adj}(\operatorname{adj} A)| = |\operatorname{adj} A|^{n-1} = (|A|^{n-1})^{n-1} = |A|^{(n-1)^2}\) Match: (IV) (C) \(\operatorname{adj}(\operatorname{adj} A ) = |A|^{n-2} A\) Match:…
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