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CUET · MATHS · PYQ PAPER 2025

Match List-l with List-ll
List-l (Matrix A)List-ll (Determinant of adjA)
(A) \(\left[\begin{array}{cc}2 & 1 \\ -1 & 2\end{array}\right]\)(I) 3
(B) \(\left[\begin{array}{ll}3 & 4 \\ 3 & 6\end{array}\right]\)(II) 6
(C) \(\left[\begin{array}{cc}3 & 7 \\ -2 & -4\end{array}\right]\)(III) 5
(D) \(\left[\begin{array}{ll}4 & 3 \\ 3 & 3\end{array}\right]\)(IV) 2
Choose the correct answer from the options given below :

  1. A (А) - (II), (В) - (I), (C) - (III), (D) - (IV)
  2. B (А) - (I), (B) - (II), (C) - (III), (D) - (IV)
  3. C (А) - (I), (B) - (II), (C) - (IV), (D) - (III)
  4. D (А) - (ІІІ), (В) - (II), (C) - (IV), (D) - (I)
Verified Solution

Answer & Solution

Correct Answer

(D) (А) - (ІІІ), (В) - (II), (C) - (IV), (D) - (I)

Step-by-step Solution

Detailed explanation

For a 2x2 matrix \(A\), \(|\text{adj}A| = |A|\). (A) \(|A| = (2)(2) - (1)(-1) = 4 + 1 = 5\). (B) \(|A| = (3)(6) - (4)(3) = 18 - 12 = 6\). (C) \(|A| = (3)(-4) - (7)(-2) = -12 + 14 = 2\). (D) \(|A| = (4)(3) - (3)(3) = 12 - 9 = 3\). Therefore: (A)-(III), (B)-(II), (C)-(IV), (D)-(I).