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To solve the given matrix equation for \( X \), let's break it down step by step:

The equation is:

\[
\begin{bmatrix} 3 & 9 \\ -1 & -8 \end{bmatrix} X + \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} X
\]

1. First, move all the terms involving \( X \) to one side. Subtract \(\begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} X\) from both sides:

\[
\left( \begin{bmatrix} 3 & 9 \\ -1 & -8 \end{bmatrix} X - \begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} X \right) = - \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix}
\]

2. Factor out \( X \) on the left-hand side:

\[
\left( \begin{bmatrix} 3 & 9 \\ -1 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} \right) X = - \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix}
\]

3. Simplify the matrix subtraction:

\[
\begin{bmatrix} 3 & 9 \\ -1 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 3-4 & 9-3 \\ -1-(-1) & -8-1 \end{bmatrix} = \begin{bmatrix} -1 & 6 \\ 0 & -9 \end{bmatrix}
\]

Now the equation is:

\[
\begin{bmatrix} -1 & 6 \\ 0 & -9 \end{bmatrix} X = - \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix}
\]

4. Multiply both sides by the inverse of \(\begin{bmatrix} -1 & 6 \\ 0 & -9 \end{bmatrix}\) to isolate \( X \).

The inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:

\[
\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]

For \( \begin{bmatrix} -1 & 6 \\ 0 & -9 \end{bmatrix} \):

\[
ad - bc = (-1)(-9) - (6)(0) = 9
\]

Thus, the inverse is:

\[
\frac{1}{9} \begin{bmatrix} -9 & -6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -\frac{2}{3} \\ 0 & -\frac{1}{9} \end{bmatrix}
\]

5. Multiply this inverse matrix by \(- \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix}\):

\[
X = \begin{bmatrix} -1 & -\frac{2}{3} \\ 0 & -\frac{1}{9} \end{bmatrix} \begin{bmatrix} 2 & -9 \\ 5 & 8 \end{bmatrix}
\]

Now, perform the matrix multiplication:

First row, first column:

\[
(-1)(2) + \left(-\frac{2}{3}\right)(5) = -2 - \frac{10}{3} = -\frac{16}{3}
\]

First row, second column:

\[
(-1)(-9) + \left(-\frac{2}{3}\right)(8) = 9 - \frac{16}{3} = \frac{27}{3} - \frac{16}{3} = \frac{11}{3}
\]

Second row, first column:

\[
0(2) + \left(-\frac{1}{9}\right)(5) = -\frac{5}{9}
\]

Second row, second column:

\[
0(-9) + \left(-\frac{1}{9}\right)(8) = -\frac{8}{9}
\]

Thus, \( X \) is:

\[
X = \begin{bmatrix} -\frac{16}{3} & \frac{11}{3} \\ -\frac{5}{9} & -\frac{8}{9} \end{bmatrix}
\]

Do you have any questions or would you like more details on any step?

Here are 5 related questions for further exploration:
1. How do you verify if a matrix has an inverse?
2. What are the conditions for matrix multiplication to be defined?
3. How do you calculate the determinant of a 3x3 matrix?
4. How can you apply Cramer's Rule to solve a system of equations?
5. What is the geometric interpretation of the inverse of a matrix?

**Tip**: Always check the determinant before attempting to find the inverse of a matrix—if it's zero, the matrix is not invertible.

Solve for \( X \) in the matrix equation:
\[ \begin{bmatrix} 3 & 9 \\ -1 & -8 \end{bmatrix} X + \begin{bmatrix} -2 & 9 \\ -5 & -8 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -1 & 1 \end{bmatrix} X \]

Explore how to solve a linear matrix equation involving a 2x2 matrix and matrix inverses. This problem focuses on concepts of matrix algebra, including matrix subtraction and inverse matrices, with a step-by-step solution for finding the unknown matrix X. Suitable for high school and early college students.

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