The problem provided is about matrices. From the image, the matrices $A$ and $B$ are defined as:

$A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}$

The task involves matrix $A^T$, which is the transpose of matrix $A$, and an equation involving $AX - B + A^T = 0$, where the goal is to find the determinant of matrix $X$.

Steps to solve:

1. Transpose of Matrix $A$: The transpose $A^T$ is found by swapping rows and columns of $A$: $A^T = \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix}$

2. Equation Setup: The equation is $AX - B + A^T = 0$. Rearranging this to isolate $X$, we get: $AX = B - A^T$ Hence, we first need to compute $B - A^T$.

3. Matrix Subtraction $B - A^T$: Perform element-wise subtraction:

   = \begin{pmatrix} 3 - 1 & -2 - 3 \\ 1 - 2 & 4 - 5 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -1 & -1 \end{pmatrix}$$

4. Solve for $X$: Now, $AX = \begin{pmatrix} 2 & -5 \\ -1 & -1 \end{pmatrix}$. To find $X$, we need to multiply both sides of the equation by the inverse of $A$, i.e., $X = A^{-1} \left( \begin{pmatrix} 2 & -5 \\ -1 & -1 \end{pmatrix} \right)$

   First, calculate the inverse of $A$.

5. Inverse of Matrix $A$: The inverse of a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by: $A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ For $A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}$: $\text{Det}(A) = (1)(5) - (2)(3) = 5 - 6 = -1$ Therefore, the inverse of $A$ is: $A^{-1} = \frac{1}{-1} \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix}$

6. Multiplying to find $X$: Now, multiply $A^{-1}$ with $B - A^T$: $X = \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 2 & -5 \\ -1 & -1 \end{pmatrix}$ Perform the matrix multiplication: $X = \begin{pmatrix} (-5)(2) + (2)(-1) & (-5)(-5) + (2)(-1) \\ (3)(2) + (-1)(-1) & (3)(-5) + (-1)(-1) \end{pmatrix}$ $X = \begin{pmatrix} -10 - 2 & 25 - 2 \\ 6 + 1 & -15 + 1 \end{pmatrix} = \begin{pmatrix} -12 & 23 \\ 7 & -14 \end{pmatrix}$

7. Determinant of $X$: Finally, the determinant of $X$ is: $\text{Det}(X) = (-12)(-14) - (23)(7) = 168 - 161 = 7$

Thus, the determinant of matrix $X$ is 7.

Would you like more details on any step or further explanation?

Here are 5 related questions to expand your understanding:

1. How do you calculate the determinant of a 3x3 matrix?
2. What properties of determinants are useful when multiplying matrices?
3. Can the determinant of a matrix be zero? If so, what does that imply about the matrix?
4. How do you solve matrix equations involving inverses if the matrix is singular?
5. What is the geometric interpretation of the determinant of a matrix?

Tip: The determinant of a matrix not only helps in solving equations but also provides insight into whether the matrix is invertible or not (non-zero determinant means invertible).