Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right]=\left[\begin{array}{cc}\log _5 128 & \log _4 5 \ \log _5 8 & \log _4 25\end{array}\right]$. If $\mathrm{A}_{i j}$ is the cofactor of \(\mathrm{a}_{i j}, \mathrm{C}_{i j}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{i \mathrm{k}} \mathrm{A}_{j \mathrm{k}}, 1…

Question

Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right]=\left[\begin{array}{cc}\log _5 128 & \log _4 5 \ \log _5 8 & \log _4 25\end{array}\right]$. If $\mathrm{A}_{i j}$ is the cofactor of $\mathrm{a}_{i j}, \mathrm{C}_{i j}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{i \mathrm{k}} \mathrm{A}_{j \mathrm{k}}, 1 \leq i, j \leq 2$, and $\mathrm{C}=\left[\mathrm{C}_{i j}\right]$, then $8|\mathrm{C}|$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} & |\mathrm{A}|=\frac{11}{2} \ & \mathrm{C}_{11}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{12} \mathrm{~A}_{12}=|\mathrm{A}|=\frac{11}{2} \ & \mathrm{C}_{12}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=0 \ & \mathrm{C}_{21}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=0 \ & \mathrm{C}_{22}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=|\mathrm{A}|=\frac{11}{2} \ & \mathrm{C}=\left[\begin{array}{cc}11 / 2 & 0 \ 0 & 11 / 2\end{array}ight] \ & |\mathrm{C}|=\frac{121}{4} \ & 8|\mathrm{C}|=242\end{aligned}$

Let \(\mathrm{A}=\left[\mathrm{a}_{i j}\right]=\left[\begin{array}{cc}\log _5 128 & \log _4 5 \\ \log _5 8 &…

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