A = [3 6 4 x] B = [5 -2  3 4]  find det(5A^-1,B^t) = 5

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.

Discover how to find the value of x in the matrix equation det(5A^-1, B^t) = 5. This problem covers matrix inversion, determinants, and matrix transposition, with a detailed step-by-step explanation. Suitable for students in Grades 10-12.

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