We are given the following matrices:

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

And we need to find $\det(5A^{-1}B^T) = 5$, where $B^T$ is the transpose of $B$.

Step 1: Finding the Determinant of $A$

The determinant of matrix $A$ is:

$\det(A) = (3 \times x) - (6 \times 4) = 3x - 24$

Step 2: Determinant of the Transpose of $B$

The determinant of $B^T$ is the same as the determinant of $B$, and the determinant of $B$ is calculated as follows:

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

Step 3: Using the Determinant Formula

We are asked to find $\det(5A^{-1}B^T)$. Using properties of determinants, we can break this down as:

$\det(5A^{-1}B^T) = \det(5) \cdot \det(A^{-1}) \cdot \det(B^T)$

Since $\det(B^T) = \det(B) = 26$, and $\det(A^{-1}) = \frac{1}{\det(A)}$, we get:

$\det(5A^{-1}B^T) = 5^2 \cdot \frac{1}{\det(A)} \cdot \det(B)$ $= 25 \cdot \frac{1}{\det(A)} \cdot 26$

Step 4: Setting the Determinant Equal to 5

We are given that $\det(5A^{-1}B^T) = 5$, so:

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

Solving for $\det(A)$:

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

Step 5: Solving for $x$

We found earlier that $\det(A) = 3x - 24$, so:

$3x - 24 = 130$ $3x = 154$ $x = \frac{154}{3} = 51.33$

Thus, the value of $x$ is approximately $51.33$.

Would you like more details, or do you have any further questions?

Here are 5 related questions to expand this problem:

1. How do you find the inverse of a 2x2 matrix?
2. What is the significance of matrix transposition in determinant calculations?
3. How do scalar multiplication properties affect matrix determinants?
4. Can you explain how the determinant helps determine the invertibility of a matrix?
5. What is the geometric interpretation of a determinant in two dimensions?

Tip: For larger matrices, calculating the determinant can be significantly more complex and may require advanced methods like cofactor expansion or row reduction.