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The question involves solving for the matrix \( X \) in the equation:

\[
\begin{pmatrix} 3 & 2 \\ -5 & -3 \end{pmatrix} \cdot X = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.
\]

### Step-by-Step Solution:

1. **Rewrite the equation**:
   The equation is:
   \[
   A \cdot X = B,
   \]
   where:
   \[
   A = \begin{pmatrix} 3 & 2 \\ -5 & -3 \end{pmatrix}, \quad
   B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.
   \]

   To find \( X \), multiply both sides by \( A^{-1} \) (the inverse of \( A \)) from the left:
   \[
   X = A^{-1} \cdot B.
   \]

2. **Find the inverse of \( A \)**:
   The inverse of a \( 2 \times 2 \) matrix is calculated as:
   \[
   A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A),
   \]
   where:
   - \(\text{det}(A) = (3)(-3) - (2)(-5) = -9 + 10 = 1\),
   - \(\text{adj}(A) = \begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix}\) (swap the diagonal elements and change the sign of the off-diagonal elements).

   Thus:
   \[
   A^{-1} = \frac{1}{1} \cdot \begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix}
   = \begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix}.
   \]

3. **Calculate \( X \)**:
   Multiply \( A^{-1} \) by \( B \):
   \[
   X = A^{-1} \cdot B = \begin{pmatrix} -3 & -2 \\ 5 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}.
   \]

   Perform the matrix multiplication:
   \[
   X_{11} = (-3)(1) + (-2)(3) = -3 - 6 = -9, \quad
   X_{12} = (-3)(2) + (-2)(4) = -6 - 8 = -14,
   \]
   \[
   X_{21} = (5)(1) + (3)(3) = 5 + 9 = 14, \quad
   X_{22} = (5)(2) + (3)(4) = 10 + 12 = 22.
   \]

   Therefore:
   \[
   X = \begin{pmatrix} -9 & -14 \\ 14 & 22 \end{pmatrix}.
   \]

### Final Answer:
\[
X = \begin{pmatrix} -9 & -14 \\ 14 & 22 \end{pmatrix}.
\]

Let me know if you want further explanation or a verification step!

### Related Questions:
1. How do you calculate the determinant of a matrix?
2. What is the adjugate matrix and how is it used in finding inverses?
3. How do you verify if a matrix is invertible?
4. Can this process be extended to larger matrices, and if so, how?
5. What are the practical applications of solving matrix equations?

**Tip**: Always check the determinant of a matrix before attempting to find its inverse; a determinant of zero means the matrix is not invertible.

Solve for the matrix X in the equation: \( \begin{pmatrix} 3 & 2 \\ -5 & -3 \end{pmatrix} \cdot X = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \).

Learn how to solve the matrix equation A \cdot X = B, where A = \begin{pmatrix} 3 & 2 \\ -5 & -3 \end{pmatrix} and B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. This detailed step-by-step solution covers matrix inverses and multiplication, suitable for high school students in Grades 11-12.

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