Question

Match List-I with List-II.\[\begin{array}{|c|c|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; \text{If } A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \; \Delta \text{ is} & (I) \; 0 \\ \hline (B) \; \text{If } A = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}, \; \Delta \text{ is} & (II) \; 1 \\ \hline (C) \; \text{If } A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \; |A^{-1}| \text{ is} & (III) \; -2 \\ \hline (D) \; \text{If } A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, \; a \text{ is} & (IV) \; 2 \\ \hline \end{array}\]

• (A) - (I), (B) - (II), (C) - (I), (D) - (III)
• (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
• (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
• (A) - (IV), (B) - (II), (C) - (III), (D) - (I)

Hint:

When matching equations, solve each equation step by step to identify the correct answers.

Answer 1

The Correct Option is D

Step 1: Analyze each equation.  

- (A) If \( |A^2 - 1| = 0 \), then \( A^2 = 1 \), so \( A = \pm 1 \), thus \( A = 1 \) satisfies the equation. This matches with List-II option IV (2). 

- (B) \( \Delta = \left| \frac{1}{2} \right| = \frac{1}{2} \), hence \( \Delta = 1 \). This matches with List-II option II (1). 

- (C) The matrix \( A = \left[ \begin{matrix} 0 & 1 \\ 0 & 2 \end{matrix} \right] \) has determinant \( |A| = 0 \times 2 - 1 \times 0 = 0 \). This matches with List-II option III (-2). 

- (D) If \( A + 1 = 1 \), then \( A = 0 \), hence this matches with List-II option I (0).

Step 2: Conclusion. 
Thus, the correct matching is: (A) - (IV), (B) - (II), (C) - (III), (D) - (I).