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The problem provided asks to find the determinant of matrix \( X \), given the matrices \( A \) and \( B \), and the equation \( AX = B + A^T \), where \( A^T \) is the transpose of matrix \( A \).

Let's break it down:

### Given Matrices:
\[
A = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -3 & -1 \\ -17 & 0 \end{bmatrix}
\]
The equation is:
\[
AX = B + A^T
\]
where \( A^T \) is the transpose of \( A \), so:
\[
A^T = \begin{bmatrix} 3 & 0 \\ 2 & 5 \end{bmatrix}
\]
Now, the equation becomes:
\[
AX = B + A^T = \begin{bmatrix} -3 & -1 \\ -17 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 2 & 5 \end{bmatrix}
\]
\[
B + A^T = \begin{bmatrix} -3 + 3 & -1 + 0 \\ -17 + 2 & 0 + 5 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -15 & 5 \end{bmatrix}
\]
So the equation simplifies to:
\[
AX = \begin{bmatrix} 0 & -1 \\ -15 & 5 \end{bmatrix}
\]
To find \( X \), we need to multiply both sides of the equation by the inverse of matrix \( A \):
\[
X = A^{-1} \begin{bmatrix} 0 & -1 \\ -15 & 5 \end{bmatrix}
\]

### Step 1: Find \( A^{-1} \)
The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
First, calculate the determinant of \( A \):
\[
\text{det}(A) = (3)(5) - (0)(2) = 15
\]
Now, the inverse of \( A \) is:
\[
A^{-1} = \frac{1}{15} \begin{bmatrix} 5 & -2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} \frac{5}{15} & \frac{-2}{15} \\ 0 & \frac{3}{15} \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{-2}{15} \\ 0 & \frac{1}{5} \end{bmatrix}
\]

### Step 2: Multiply \( A^{-1} \) with \( \begin{bmatrix} 0 & -1 \\ -15 & 5 \end{bmatrix} \)
\[
X = \begin{bmatrix} \frac{1}{3} & \frac{-2}{15} \\ 0 & \frac{1}{5} \end{bmatrix} \begin{bmatrix} 0 & -1 \\ -15 & 5 \end{bmatrix}
\]
Now perform the matrix multiplication:
\[
X_{11} = \left(\frac{1}{3} \times 0\right) + \left(\frac{-2}{15} \times -15\right) = 0 + 2 = 2
\]
\[
X_{12} = \left(\frac{1}{3} \times -1\right) + \left(\frac{-2}{15} \times 5\right) = -\frac{1}{3} + \left(\frac{-10}{15}\right) = -\frac{1}{3} - \frac{2}{3} = -1
\]
\[
X_{21} = \left(0 \times 0\right) + \left(\frac{1}{5} \times -15\right) = 0 - 3 = -3
\]
\[
X_{22} = \left(0 \times -1\right) + \left(\frac{1}{5} \times 5\right) = 0 + 1 = 1
\]
Thus, matrix \( X \) is:
\[
X = \begin{bmatrix} 2 & -1 \\ -3 & 1 \end{bmatrix}
\]

### Step 3: Find the determinant of \( X \)
The determinant of matrix \( X \) is \( -1 \).

Would you like more details or have any further questions? Here are some related problems for practice:

1. Calculate the determinant of a matrix using different methods.
2. Solve a similar problem but with different matrices \( A \) and \( B \).
3. How do you find the inverse of a 3x3 matrix?
4. Can a matrix with a determinant of zero have an inverse?
5. What are the properties of determinants in matrix multiplication?

**Tip:** When multiplying matrices, make sure to check the dimensions of the matrices to ensure they can be multiplied. The number of columns in the first matrix must equal the number of rows in the second matrix.

Diketahui matriks A = [3 2; 0 5] dan B = [-3 -1; -17 0]. Jika AX = B + A^T, tentukan determinan matriks X!

Solve the matrix equation AX = B + A^T where A and B are given matrices, and determine the determinant of matrix X. This problem includes matrix multiplication, matrix inversion, and determinant calculation, suitable for high school students studying advanced algebra.

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