The given equation is:

$B^{-1} X A^{-1} = C^{-1}$

with the matrices:

$A = \begin{pmatrix} -1 & -3 \\ 2 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} -2 & 4 \\ 0 & 6 \end{pmatrix}, \quad C = \begin{pmatrix} 2 & 0 \\ 10 & -1 \end{pmatrix}.$

The task is to find $X$.

Steps:

1. First, find the inverses $B^{-1}, A^{-1}, C^{-1}$.
2. Then solve for $X$ by rearranging the equation: $X = B C^{-1} A$ (multiplying both sides by $B$ and $A$).

I will calculate these now.Here are the results of the calculations:

1. The inverse of matrix $A$ is: $A^{-1} = \begin{pmatrix} 0 & 0.5 \\ -0.3333 & -0.1667 \end{pmatrix}$

2. The inverse of matrix $B$ is: $B^{-1} = \begin{pmatrix} -0.5 & 0.3333 \\ 0 & 0.1667 \end{pmatrix}$

3. The inverse of matrix $C$ is: $C^{-1} = \begin{pmatrix} 0.5 & 0 \\ 5 & -1 \end{pmatrix}$

4. The matrix $X$ is: $X = \begin{pmatrix} -27 & -57 \\ -42 & -90 \end{pmatrix}$

Would you like more details or explanations on any step?

Here are 5 questions related to this problem:

1. What are the properties of matrix inverses?
2. How do you compute the determinant of a 2x2 matrix?
3. Can every matrix be inverted? Why or why not?
4. What is the significance of the inverse of a matrix in linear algebra?
5. How would the solution change if matrix $B$ were not invertible?

Tip: To verify if a matrix is invertible, its determinant should not be zero.