To find the value of $x$ such that the determinant of $5A^{-1}$ and $B^t$ equals 5, we need to break this down step-by-step:

Given matrices:

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

And the equation is:

$\text{det}(5A^{-1}, B^t) = 5$

Step 1: Finding $A^{-1}$

First, let's compute the inverse of $A$. The formula for the inverse of a 2x2 matrix is:

$A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

where the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

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

• $a = 3$
• $b = 6$
• $c = 4$
• $d = x$

The determinant of $A$ is:

$\text{det}(A) = (3)(x) - (6)(4) = 3x - 24$

Thus,

$A^{-1} = \frac{1}{3x - 24} \begin{bmatrix} x & -6 \\ -4 & 3 \end{bmatrix}$

Step 2: Finding $5A^{-1}$

Now, multiply $A^{-1}$ by 5:

$5A^{-1} = \frac{5}{3x - 24} \begin{bmatrix} x & -6 \\ -4 & 3 \end{bmatrix}$

Step 3: Compute the determinant of $5A^{-1}$ and $B^t$

The next step involves finding the determinant of $5A^{-1}$ and $B^t$, where $B^t$ is the transpose of matrix $B$.

$B^t = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}$

Now, we compute the determinant of $5A^{-1}, B^t$, which means finding the determinant of the matrix:

$\text{det} \left( \frac{5}{3x - 24} \begin{bmatrix} x & -6 \\ -4 & 3 \end{bmatrix}, \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix} \right)$

To match the condition that $\text{det}(5A^{-1}, B^t) = 5$, we can solve for $x$.

Let me calculate this determinant next.