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

The expression for $$C_{jk}$$ is the definition of a determinant if $$j=i$$, but here the indices are specific. Note that $$C = A^T \cdot \text{adj}(A)^T$$. However, a simpler property of cofactors is that $$A \cdot \text{adj}(A) = |A|I$$.

Looking at the sum $$C_{jk} = \sum a_{ik}A_{ik}$$, this represents the product of elements of a column and cofactors of a column.

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

Using base change: $$\log_b a = \frac{\ln a}{\ln b}$$

$$|A| = (\frac{7 \ln 2}{\ln 5})(\frac{2 \ln 5}{2 \ln 2}) - (\frac{\ln 5}{2 \ln 2})(\frac{3 \ln 2}{\ln 5})$$

$$|A| = 7 - \frac{3}{2} = \frac{11}{2} = 5.5$$

The matrix $$C$$ results in $$|C| = |A|^2$$ (due to the property of the product of $$A$$ and its cofactor matrix).

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

$$8|C| = 8 \times \frac{121}{4} = 2 \times 121 = \mathbf{242}$$

Correct Option: C