The problem involves solving for matrix $X$ in the given matrix equation:

$\begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} X = \begin{bmatrix} -3 & 5 \\ -5 & 2 \end{bmatrix}$

Matrix $X$ is likely a $2 \times 2$ matrix (as indicated by the empty boxes), and we need to solve for it. Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, and substitute into the equation to get:

$\begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -3 & 5 \\ -5 & 2 \end{bmatrix}$

Performing the matrix multiplication on the left-hand side:

$\begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a - 6c & b - 6d \\ c & d \end{bmatrix}$

Now, set this equal to the matrix on the right-hand side:

$\begin{bmatrix} a - 6c & b - 6d \\ c & d \end{bmatrix} = \begin{bmatrix} -3 & 5 \\ -5 & 2 \end{bmatrix}$

From this, we get the following system of equations:

1. $a - 6c = -3$
2. $b - 6d = 5$
3. $c = -5$
4. $d = 2$

Substitute $c = -5$ and $d = 2$ into the first two equations:

1. $a - 6(-5) = -3 \implies a + 30 = -3 \implies a = -33$
2. $b - 6(2) = 5 \implies b - 12 = 5 \implies b = 17$

Thus, the matrix $X$ is:

$X = \begin{bmatrix} -33 & 17 \\ -5 & 2 \end{bmatrix}$

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. What is the process for finding the inverse of a matrix?
2. How would you solve the system if the matrix equation had more than two variables?
3. Can the original matrix $\begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix}$ be classified as an upper triangular matrix?
4. How does matrix multiplication differ from scalar multiplication?
5. How can you verify that the solution is correct by substituting it back into the equation?

Tip: Always double-check your matrix multiplications by performing each step carefully to avoid errors in sign or value.