Solve the matrix equation $A X=B$ for $X$.
\[
A=\left[\begin{array}{rr}
1 & 4 \\
-6 & 4
\end{array}\right], B=\left[\begin{array}{r}
-12 \\
-40
\end{array}\right]
\]
\[
x=\left[\begin{array}{l}
\square \\
\square
\end{array}\right]
\]

Step 1 ：We are given the matrix equation $AX = B$, where $A = \left[\begin{array}{rr} 1 & 4 \\ -6 & 4 \end{array}\right]$ and $B = \left[\begin{array}{r} -12 \\ -40 \end{array}\right]$. We are asked to solve for $X$.

Step 2 ：The matrix equation $AX = B$ can be solved for $X$ by multiplying both sides of the equation by the inverse of $A$, if it exists. The inverse of a matrix $A$ is denoted $A^{-1}$ and it has the property that $AA^{-1} = A^{-1}A = I$, where $I$ is the identity matrix.

Step 3 ：So, if we multiply both sides of the equation by $A^{-1}$, we get $A^{-1}AX = A^{-1}B$, which simplifies to $X = A^{-1}B$. Therefore, to solve the equation, we need to find the inverse of $A$ and then multiply it by $B$.

Step 4 ：First, we find the inverse of $A$, denoted as $A_{inv} = \left[\begin{array}{rr} 0.14285714 & -0.14285714 \\ 0.21428571 & 0.03571429 \end{array}\right]$.

Step 5 ：Then, we multiply $A_{inv}$ by $B$ to get $X = \left[\begin{array}{r} 4 \\ -4 \end{array}\right]$.

Step 6 ：\(\boxed{X = \left[\begin{array}{r} 4 \\ -4 \end{array}\right]}\) is the solution to the matrix equation.