QJEEM25Maf75b: JEE Main 2025 Mathematics - Class 12 JEEM25Maf75b Solution: Let . If is the cofactor of , , and , then is equal to :

To solve the problem, we need to determine the determinant of matrix $A$ :

$A = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}$

The determinant of $A$ , denoted as $|A|$ , is calculated as:

$|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)$

Evaluating each component:

$\log_5 128$ can be simplified using change of base formula:

$\log_5 128 = \frac{\log_{10}128}{\log_{10}5}$

$\log_4 5$ using change of base:

$\log_4 5 = \frac{\log_{10}5}{\log_{10}4}$

$\log_5 8$ :

$\log_5 8 = \frac{\log_{10}8}{\log_{10}5}$

$\log_4 25$ :

$\log_4 25 = \frac{\log_{10}25}{\log_{10}4}$

Now, substitute these values into $|A|$ :

$|A| = \left(\frac{\log_{10}128}{\log_{10}5}\right) \left(\frac{\log_{10}25}{\log_{10}4}\right) - \left(\frac{\log_{10}5}{\log_{10}4}\right) \left(\frac{\log_{10}8}{\log_{10}5}\right)$

Next, we find the cofactors of matrix $A$ :

$A_{11} = \log_4 25$

$A_{12} = -\log_5 8$

$A_{21} = -\log_4 5$

$A_{22} = \log_5 128$

Then, calculate matrix $C$ whose elements are given by $C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk}$ :

$C_{11} = a_{11}A_{11} + a_{12}A_{12} = |A| = \frac{11}{2}$

$C_{12} = a_{11}A_{21} + a_{12}A_{22} = 0$

$C_{21} = a_{21}A_{11} + a_{22}A_{12} = 0$

$C_{22} = a_{21}A_{21} + a_{22}A_{22} = |A| = \frac{11}{2}$

Thus, matrix $C$ is:

$C = \begin{bmatrix} \frac{11}{2} & 0 \\ 0 & \frac{11}{2} \end{bmatrix}$

To find $|C|$ , we compute:

$|C| = \left(\frac{11}{2}\right)\left(\frac{11}{2}\right) = \frac{121}{4}$

Finally, calculate $8|C|$ :

$8|C| = 8 \times \frac{121}{4} = 242$