no. there is matrix * X = some other matrix. look once again

Ah, I see what you're referring to! Upon closer inspection, the problem involves solving for the determinant of matrix \( X \), and the equation provided is:

\[
A \cdot X = B,
\]

where \( A = \begin{pmatrix} 3 & 1 \\ 4 & -2 \end{pmatrix} \) and \( B = \begin{pmatrix} -1 & -1 \\ 3 & 3 \end{pmatrix} \).

### Step-by-step solution:

1. **Interpretation:**
   To solve for \( X \), we use the property:
   \[
   X = A^{-1} \cdot B,
   \]
   where \( A^{-1} \) is the inverse of matrix \( A \).

2. **Find the determinant of \( A \):**
   The determinant of \( A \) is given by:
   \[
   \text{det}(A) = (3)(-2) - (1)(4) = -6 - 4 = -10.
   \]

3. **Find the inverse of \( A \):**
   The formula for the inverse of a \( 2 \times 2 \) matrix is:
   \[
   A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix},
   \]
   where \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).

   Substituting the values from \( A = \begin{pmatrix} 3 & 1 \\ 4 & -2 \end{pmatrix} \):
   \[
   A^{-1} = \frac{1}{-10} \begin{pmatrix} -2 & -1 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} 0.2 & 0.1 \\ 0.4 & -0.3 \end{pmatrix}.
   \]

4. **Calculate \( X = A^{-1} \cdot B \):**
   Multiply \( A^{-1} \) by \( B = \begin{pmatrix} -1 & -1 \\ 3 & 3 \end{pmatrix} \):
   \[
   X = \begin{pmatrix} 0.2 & 0.1 \\ 0.4 & -0.3 \end{pmatrix} \cdot \begin{pmatrix} -1 & -1 \\ 3 & 3 \end{pmatrix}.
   \]

   Perform the matrix multiplication:
   \[
   X_{11} = (0.2)(-1) + (0.1)(3

Discover how to solve for matrix X in the equation A * X = B. This detailed step-by-step solution includes finding the determinant and inverse of matrix A and performing matrix multiplication to compute X. Key concepts include matrix inversion and determinant calculation, suitable for advanced high school or college students.

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