We are given two matrices $A$ and $B$ as follows:

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

The task is to find the determinant of $5A^{-1} B^t$, where the determinant is equal to 5.

Step 1: Using properties of determinants

We know the following property of determinants: $\text{det}(k A) = k^n \cdot \text{det}(A)$ where $k$ is a scalar and $A$ is an $n \times n$ matrix. Here, the matrix $A$ is $2 \times 2$, so: $\text{det}(5A^{-1} B^t) = 5^2 \cdot \text{det}(A^{-1}) \cdot \text{det}(B^t)$ Since the determinant of the transpose of a matrix is the same as the determinant of the matrix itself: $\text{det}(5A^{-1} B^t) = 25 \cdot \text{det}(A^{-1}) \cdot \text{det}(B)$

Also, we know that: $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}$ Thus, the equation becomes: $\text{det}(5A^{-1} B^t) = 25 \cdot \frac{1}{\text{det}(A)} \cdot \text{det}(B)$

Step 2: Determinant of $B$

First, we calculate the determinant of matrix $B$: $\text{det}(B) = (5)(4) - (-2)(3) = 20 + 6 = 26$

Step 3: Determinant of $A$

We don't yet know $x$, so we calculate the determinant of matrix $A$ symbolically: $\text{det}(A) = (3)(x) - (6)(4) = 3x - 24$

Step 4: Substituting into the determinant equation

Substitute the determinant values into the equation: $5 = 25 \cdot \frac{26}{3x - 24}$ Solve for $x$: $5 = \frac{650}{3x - 24}$ Multiplying both sides by $3x - 24$ to eliminate the fraction: $5(3x - 24) = 650$ Expanding the left-hand side: $15x - 120 = 650$ Add 120 to both sides: $15x = 770$ Finally, divide by 15: $x = \frac{770}{15} = 51.33$

Thus, $x = \frac{770}{15}$ or approximately $x \approx 51.33$.

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

5 Related Questions:

1. How do you calculate the inverse of a 2x2 matrix?
2. What are the properties of matrix transposition in determinants?
3. Can the determinant of a matrix ever be negative, and what does it imply?
4. How can we find the eigenvalues of matrix $A$ or $B$?
5. Why is the determinant of $A^{-1}$ the reciprocal of the determinant of $A$?

Tip: When solving matrix equations involving determinants, using properties like $\det(A^{-1}) = \frac{1}{\det(A)}$ can simplify calculations significantly.