Animated Solution for Mathematics - Matrices and Determinants: Let $A=[a_{ij}]=\left[\begin{array}{cc}{\log }_{5}128& {\log }_{4}5\\ {\log }_{5}8& {\log }_{4}25\end{array}\right].$ If $A_{ij}$ is the cofactor of $a_{ij}$, $C_{ij}={\sum }_{k=1}^2{a}_{ik}{A}_{jk},1\le i,j\le 2$, and $C=[C_{ij}]$, then $8|C|$ is equal to:

Welcome, fellow traveler in the realm of mathematics. Today, we are going to dissect a problem that, at first glance, looks like a tedious exercise in logarithmic arithmetic.

You see a matrix $A=\left[\begin{array}{cc}{\log }_{5}128& {\log }_{4}5\\ {\log }_{5}8& {\log }_{4}25\end{array}\right]$ and your instinct might be to dive straight into calculating cofactors. But wait! Before you sharpen your pencil, let us pause and look for the hidden architecture.

The problem defines $C_{ij}={\sum }_{k=1}^2{a}_{ik}{A}_{jk}$. If you have spent time with the properties of determinants, this expression should make your heart skip a beat.

This is the definition of the product of a matrix and its adjugate. We know that for any square matrix $A$, the sum of the products of elements of row $i$ with the cofactors of row $j$ is given by the property:

where ${\delta }_{ij}$ is $1$ if $i=j$ and $0$ if $i eq j$. This means our matrix $C$ is not a chaotic collection of numbers; it is a diagonal matrix: $C=\left[\begin{array}{cc}|A|& 0\\ 0& |A|\end{array}\right]$.

Consequently, the determinant $|C|$ is simply $|A{|}^2$. We have just reduced a potentially grueling task into a simple calculation of the determinant of $A$.

Now, let us tackle $|A|=\left|\begin{array}{cc}{\log }_{5}128& {\log }_{4}5\\ {\log }_{5}8& {\log }_{4}25\end{array}\right|$. Using the power rule ${\log }_b{a}^n=n{\log }_{b}a$, we can simplify the entries:

Substituting these into our determinant expression, we get:

Notice the common factor $({\log }_{5}2\cdot {\log }_{4}5)$? Factoring it out is like finding the golden key to the lock:

Finally, we apply the change of base formula, ${\log }_{b}a=\frac{\ln a}{\ln b}$. The product becomes:

Since $\ln 4=\ln 2^2=2\ln 2$, the expression simplifies to $\frac{\ln 2}{2\ln 2}=\frac{1}{2}$. Thus, $|A|=11\times \frac{1}{2}=\frac{11}{2}$.

Returning to our earlier discovery, $|C|=|A{|}^2=(\frac{11}{2}{)}^2=\frac{121}{4}$. The question asks for $8|C|$, so:

And there it is! Through the power of matrix properties and the elegance of logarithmic identities, we have arrived at the final answer of 242. Never fear the complexity of a problem; look for the structure, trust the properties, and the math will reveal its beauty to you.