Let . If is the cofactor of , and , then 8|C| is equal to : [2025]

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Solution

Q.
Let $A = [a_{i j}] = [\begin{matrix} \log_{5} 128 & \log_{4} 5 \\ \log_{5} 8 & \log_{4} 25 \end{matrix}]$ . If $A_{i j}$ is the cofactor of $a_{i j}, C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$ , and $C = [C_{i j}]$ , then $8$ $| C |$ is equal to : [2025]

1 242

2 288

3 262

4 222

Ans.
(1)

From the given matrix A, $| A |$ $= \frac{11}{2}$

Here, $C_{i j} = \sum_{k = 1}^{2} a_{i k} A_{j k}, 1 \leq i, j \leq 2$

$C_{11} = \sum_{k = 1}^{2} a_{1 k} A_{1 k} = a_{11} A_{11} + a_{12} A_{12} = | A | = \frac{11}{2}$

$C_{12} = \sum_{k = 1}^{2} a_{1 k} A_{2 k} = 0$

$C_{21} = \sum_{k = 1}^{2} a_{2 k} A_{1 k} = 0$

$C_{22} = \sum_{k = 1}^{2} a_{2 k} A_{2 k} = | A | = \frac{11}{2}$

Now, $C = [\begin{matrix} 11 / 2 & 0 \\ 0 & 11 / 2 \end{matrix}] \Rightarrow$ $| C |$ $= \frac{121}{4}$

Hence, $8$ $| C |$ $= 242$ .

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