Let A= [a_ i j ]= [ cc _ 5 128 & _ 4 5 _ 5 8 & _ 4 25 ]. If A_ ij… - Vidyadip

MCQ

Let $A=\left[a_{i j}\right]=\left[\begin{array}{cc}\log _{5} 128 & \log _{4} 5 \\ \log _{5} 8 & \log _{4} 25\end{array}\right]$. If $\mathrm{A}_{\mathrm{ij}}$ is the cofactor of $\mathrm{a}_{\mathrm{ij}}, \mathrm{C}_{\mathrm{ij}}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{\mathrm{ik}} \mathrm{A}_{\mathrm{jk}}, 1 \leq \mathrm{i}$, $\mathrm{j} \leq 2$, and $\mathrm{C}=\left[\mathrm{C}_{\mathrm{ij}}\right]$, then $8|\mathrm{C}|$ is equal to :

A
262
B
288
✓
242
D
222

✓

Answer

Correct option: C.

242

(C)
$
\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}\right] \\
& |\mathrm{C}|=\frac{121}{4} \\
& 8|\mathrm{C}|=242
\end{aligned}
$

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