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Let's solve the given matrix equation for \( X \).

The equation is:
\[ \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X + \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix} = \begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix} \]

First, isolate \( X \):
\[ \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X = \begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix} - \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix} \]

Calculate the matrix on the right side:
\[ \begin{bmatrix} -9 & 1 \\ -8 & -5 \end{bmatrix} - \begin{bmatrix} 4 & -7 \\ -3 & -8 \end{bmatrix} = \begin{bmatrix} -9 - 4 & 1 + 7 \\ -8 + 3 & -5 + 8 \end{bmatrix} = \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix} \]

So the equation becomes:
\[ \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} X = \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix} \]

To solve for \( X \), multiply both sides by the inverse of the matrix \( \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} \).

First, find the inverse of \( \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} \):

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).

For \( \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} \):
\[ \text{Determinant} = (-2 \times -6) - (-6 \times -7) = 12 - 42 = -30 \]

The inverse is:
\[ \frac{1}{-30} \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \\ -\frac{7}{30} & \frac{1}{15} \end{bmatrix} \]

Now, multiply the inverse by the matrix \( \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix} \):
\[ X = \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \\ -\frac{7}{30} & \frac{1}{15} \end{bmatrix} \begin{bmatrix} -13 & 8 \\ -5 & 3 \end{bmatrix} \]

Calculate each element of the resulting matrix:

1. First element:
\[ X_{11} = \left(\frac{1}{5} \times -13\right) + \left(-\frac{1}{5} \times -5\right) = -\frac{13}{5} + \frac{5}{5} = -\frac{13}{5} + 1 = -\frac{13 - 5}{5} = -\frac{8}{5} = -\frac{8}{5} \]

2. Second element:
\[ X_{12} = \left(\frac{1}{5} \times 8\right) + \left(-\frac{1}{5} \times 3\right) = \frac{8}{5} - \frac{3}{5} = \frac{8 - 3}{5} = \frac{5}{5} = 1 \]

3. Third element:
\[ X_{21} = \left(-\frac{7}{30} \times -13\right) + \left(\frac{1}{15} \times -5\right) = \frac{91}{30} - \frac{5}{15} = \frac{91}{30} - \frac{10}{30} = \frac{81}{30} = \frac{27}{10} = 2.7 \]

4. Fourth element:
\[ X_{22} = \left(-\frac{7}{30} \times 8\right) + \left(\frac{1}{15} \times 3\right) = -\frac{56}{30} + \frac{3}{15} = -\frac{56}{30} + \frac{6}{30} = -\frac{50}{30} = -\frac{5}{3} \]

Therefore, the solution is:
\[ X = \begin{bmatrix} -\frac{8}{5} & 1 \\ 2.7 & -\frac{5}{3} \end{bmatrix} \]

Would you like more details or have any questions?

Here are 5 questions you might want to ask:
1. How can I verify the correctness of a matrix inverse?
2. Can you explain the determinant calculation in more detail?
3. How does matrix multiplication work?
4. What are some common mistakes when solving matrix equations?
5. Can you show me how to solve another matrix equation?

Tip: Always double-check your calculations, especially when working with fractions and negative signs, to avoid common arithmetic errors.

Learn how to solve a matrix equation involving the calculation of the inverse of a 2x2 matrix. This problem covers key concepts in matrix algebra. Suitable for advanced high school students.

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