The problem provided asks us to find the determinant of $5A^{-1}B^T$, given matrices $A$ and $B$:

$A = \begin{bmatrix} 3 & 6 \\ 4 & x \end{bmatrix}, \quad B = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}$

It is also given that $\det(5A^{-1}B^T) = 5$.

Step-by-Step Solution:

1. Using properties of determinants:

The determinant of a product of matrices follows the rule: $\det(5A^{-1}B^T) = \det(5) \cdot \det(A^{-1}) \cdot \det(B^T)$

Since $\det(kM) = k^n \det(M)$ for an $n \times n$ matrix $M$, and here we have $2 \times 2$ matrices, we get: $\det(5A^{-1}B^T) = 5^2 \cdot \det(A^{-1}) \cdot \det(B^T) = 25 \cdot \det(A^{-1}) \cdot \det(B^T)$

The determinant of an inverse matrix is given by: $\det(A^{-1}) = \frac{1}{\det(A)}$

Also, the determinant of the transpose of a matrix is the same as the determinant of the matrix: $\det(B^T) = \det(B)$

Thus, the equation becomes: $25 \cdot \frac{1}{\det(A)} \cdot \det(B) = 5$

2. Determinant of $B$:

$\det(B) = (5)(4) - (3)(-2) = 20 + 6 = 26$

3. Plug values into the equation:

$25 \cdot \frac{26}{\det(A)} = 5$

Simplify: $\frac{650}{\det(A)} = 5$

Solving for $\det(A)$: $\det(A) = \frac{650}{5} = 130$

4. Determinant of $A$:

The determinant of $A$ is: $\det(A) = (3)(x) - (4)(6) = 3x - 24$

From the previous step, we know: $3x - 24 = 130$

Solving for $x$: $3x = 154 \quad \Rightarrow \quad x = \frac{154}{3}$

Thus, the value of $x$ is $\frac{154}{3}$.

Would you like further details or clarifications?

Here are 5 related questions to expand the concept:

1. How is the inverse of a matrix calculated?
2. What are the properties of matrix transposes and their determinants?
3. How does the determinant of a scalar multiple of a matrix work?
4. What is the significance of matrix determinants in linear algebra?
5. Can you solve similar problems with 3x3 matrices or higher dimensions?

Tip: When dealing with inverse matrices, always remember that $\det(A^{-1}) = \frac{1}{\det(A)}$, making determinants a powerful tool for solving matrix equations.