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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.

Find det(5A^{-1}B^T) = 5, given A = [[3, 6], [4, x]] and B = [[5, -2], [3, 4]].

Learn how to solve a matrix determinant problem involving inverse matrices and transposition. This example uses matrices A and B to find x where det(5A^{-1}B^T) = 5. Explore determinant properties and matrix operations in this step-by-step guide.

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