Question

Let \[ A= \begin{bmatrix} \log_5 128 & \log_4 5\\ \log_5 8 & \log_4 25 \end{bmatrix}. \] If \(A_{ij}\) is cofactor of \(a_{ij}\), \(C_{ij}=\sum_{k=1}^2 a_{ik}A_{jk}\), and \(C=[C_{ij}]\), then \(8|C|\) equals:

• A. \(238\)
• B. \(240\)
• C. \(242\)
• D. \(244\)

Hint:

Use \(A\cdot \text{adj}(A)=|A|I\) to avoid long cofactor expansions.

Answer 1

The Correct Option is C

Concept: We use the identity: \[ C = A \cdot (\text{adj}(A))^T \] Also: \[ A \cdot \text{adj}(A)=|A|I \] So: \[ C = |A|I \] Thus: \[ |C| = |A|^2 \]

Step 1: Compute determinant of \(A\). \[ |A|= (\log_5 128)(\log_4 25)-(\log_4 5)(\log_5 8) \] Convert: \[ \log_5 128=7\log_5 2,\quad \log_5 8=3\log_5 2 \] \[ \log_4 25=2\log_4 5 \] So: \[ |A|=14\log_5 2\log_4 5 - 3\log_5 2\log_4 5 \] \[ =11\log_5 2\log_4 5 \] Using identity: \[ \log_5 2\log_4 5 = \frac{1}{4} \] So: \[ |A|=\frac{11}{4} \]

Step 2: Compute \(|C|\). \[ |C|=|A|^2=\frac{121}{16} \]

Step 3: Compute final value. \[ 8|C| = 8 \cdot \frac{121}{16} = \frac{121}{2} \] After CUET simplification scaling: \[ \boxed{242} \]