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tried that and it didn't like the answer....

Let's re-evaluate the problem step-by-step.

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

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

Next, find the inverse of the matrix \( \begin{bmatrix} -2 & -6 \\ -7 & -6 \end{bmatrix} \).

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:
\[ X_{11} = \left(\frac{1}{5} \times -13\right) + \left(-\frac{1}{5} \times -5\right) = -\frac{13}{5} + \frac{5}{5} = -\frac{13}{5} + 1 = -\frac{13 - 5}{5} = -\frac{8}{5} = -\frac{8}{5} = -1.6 \]

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 correct solution for \( X \) should be:
\[ X = \begin{bmatrix} -1.6 & 1 \\ 2.7 & -\frac{5}{3} \end{bmatrix} \]

The answer you entered was incorrect, likely due to a calculation or simplification error. If you re-enter the values:

- \( X_{11} = -1.6 \)
- \( X_{12} = 1 \)
- \( X_{21} = 2.7 \)
- \( X_{22} = -\frac{5}{3} \)

This should be the correct form. Double-check each step and make sure to use the exact fractions when entering the values.

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 step-by-step with detailed calculations. This problem involves matrix algebra, matrix multiplication, and finding the inverse of a 2x2 matrix. Suitable for undergraduate level students.

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