Algebra – Matrices – JEE Main 29 Jan 2025 Shift 1

Solution:

First, calculate the determinant of $A$, $|A|$.

$a_{11} = \log_5 128 = \log_5 (2^7) = 7\log_5 2$.

$a_{12} = \log_4 5 = \log_{2^2} 5 = \frac{1}{2}\log_2 5$.

$a_{21} = \log_5 8 = \log_5 (2^3) = 3\log_5 2$.

$a_{22} = \log_4 25 = \log_{2^2} (5^2) = \frac{2}{2}\log_2 5 = \log_2 5$.

$|A| = a_{11}a_{22} – a_{12}a_{21}$.

$|A| = (7\log_5 2)(\log_2 5) – (\frac{1}{2}\log_2 5)(3\log_5 2)$.

Using $\log_a b \cdot \log_b a = 1$:

$|A| = 7(1) – \frac{3}{2}(1) = 7 – 1.5 = 5.5 = \frac{11}{2}$.

The term $C_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}$ represents the element of the product matrix $C = A \cdot (\text{adj } A)^T$ or simply the row-cofactor expansion property.

Specifically, $\sum_{k} a_{ik} A_{jk} = |A|$ if $i=j$ and $0$ if $i \neq j$.

Thus, matrix $C = \begin{bmatrix} |A| & 0 \\ 0 & |A| \end{bmatrix} = \begin{bmatrix} \frac{11}{2} & 0 \\ 0 & \frac{11}{2} \end{bmatrix}$.

The determinant $|C| = (\frac{11}{2})^2 = \frac{121}{4}$.

We need to find $8|C|$.

$8 \times \frac{121}{4} = 2 \times 121 = 242$.

Ans. (3)