CUET · MATHS · PYQ PAPER 2025
Match List-I with List-II
| List-I ( Matrix A ) | List-II ( Determinant of adj A ) |
| (A) \(\left[\begin{array}{cc}2 & 1 \\0 & -1\end{array}\right]\) | (I) 6 |
| (B) \(\left[\begin{array}{cc}0 & 1 \\4 & -1\end{array}\right]\) | (II) 5 |
| (C) \(\left[\begin{array}{cc}1 & 2 \\-3 & -1\end{array}\right]\) | (III) \(-4\) |
| (D) \(\left[\begin{array}{cc}4 & -2 \\3 & 0\end{array}\right]\) | (IV) \(-2\) |
- A (Α) - (IV), (Β) - (III), (С) - (I), (D) - (II)
- B (А) - (I), (Β) - (II), (C) - (IV), (D) - (III)
- C (А) - (III), (В) - (IV), (C) - (I), (D) - (II)
- D (Α) - (IV), (Β) - (III), (С) - (II), (D) - (I)
Answer & Solution
Correct Answer
(D) (Α) - (IV), (Β) - (III), (С) - (II), (D) - (I)
Step-by-step Solution
Detailed explanation
For a \(2 \times 2\) matrix A, \(|\text{adj } A| = |A|\). (A) \(|A| = (2)(-1) - (1)(0) = -2\) (B) \(|A| = (0)(-1) - (1)(4) = -4\) (C) \(|A| = (1)(-1) - (2)(-3) = -1 + 6 = 5\) (D) \(|A| = (4)(0) - (-2)(3) = 0 + 6 = 6\) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
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