CUET · MATHS · PYQ PAPER 2025
Match the Matrix in List - I with its Adjoint's Determinant in List - II.
List - I (Matrix A) :
| LIST 1 | LIST 2 |
| (A) \(\left[\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right]\) | (I) 9 |
| (B) \(\left[\begin{array}{cc}5 & -1 \\ 4 & 2\end{array}\right]\) | (II) 8 |
| (C) \(\left[\begin{array}{cc}6 & -1 \\ 2 & 1\end{array}\right]\) | (III) 14 |
| (D) \(\left[\begin{array}{ll}4 & 1 \\ 3 & 3\end{array}\right]\) | (IV) 2 |
- A (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
- B (А) - (Ⅱ), (B) - (IV), (C) - (I), (D) - (III)
- C (А) - (Ⅲ), (B) - (I), (C) - (IV), (D) - (II)
- D (A) - (IV), (B) - (Ⅲ), (С) - (Ⅱ), (D) - (I)
Answer & Solution
Correct Answer
(D) (A) - (IV), (B) - (Ⅲ), (С) - (Ⅱ), (D) - (I)
Step-by-step Solution
Detailed explanation
(A) \( \left|\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right| = (3)(2) - (1)(4) = 6 - 4 = 2 \) (A) - (IV) (B) \( \left|\begin{array}{cc}5 & -1 \\ 4 & 2\end{array}\right| = (5)(2) - (-1)(4) = 10 + 4 = 14 \) (B) - (III) (C)…
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