To solve for $\det\left(5A^{-1}B^t\right) = t$, we can follow similar steps as the previous solution but focus on finding the determinant of the matrix expression.

Let's work through this step-by-step:

Step 1: Use Determinant Properties

We are tasked with finding the determinant of the matrix expression $5A^{-1}B^T$, where:

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

We can break down the determinant using properties of determinants:

• $\det(AB) = \det(A) \cdot \det(B)$
• $\det(A^T) = \det(A)$
• $\det(cA) = c^n \det(A)$ for an $n \times n$ matrix, where $c$ is a scalar (in this case, $c = 5$ and $n = 2$, as $A$ is a 2x2 matrix).

Thus, the expression can be simplified as follows:

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

Since $\det(5I) = 25$ for a 2x2 matrix, we have:

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

Step 2: Find Determinants of A and B

We already know the determinant of matrix $A$ and matrix $B$.

For ( A = \begin{bmatrix} 3 & 6 \ 4 & x \end{bmatrix} \