Solve the system of linear equations $\left\{\begin{array}{c}-6 x+2 y=-10 \\ 8 x-3 y=12\end{array}\right.$ by completing the followin
(a) Suppose the coefficient matrix is $A=\left[\begin{array}{cc}-6 & 2 \\ 8 & -3\end{array}\right]$. Find $A^{-1}$ and use it to write the solution matrix $\left[\begin{array}{l}x \\ y\end{array}\right]$ as a product of two matrices.
\[
\left[\begin{array}{l}
x \\
y
\end{array}\right]=
\]
(b) Find the solution of the system. Write your answers in simplest form.
\[
x=\square \quad y=
\]

Step 1 ：Given the system of linear equations \(\begin{cases}-6x+2y=-10\\8x-3y=12\end{cases}\), we can write this system in matrix form as \(AX=B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. In this case, \(A=\begin{bmatrix}-6 & 2\\8 & -3\end{bmatrix}\), \(X=\begin{bmatrix}x\\y\end{bmatrix}\), and \(B=\begin{bmatrix}-10\\12\end{bmatrix}\).

Step 2 ：The inverse of a 2x2 matrix \(\begin{bmatrix}a & b\\c & d\end{bmatrix}\) is given by \(\frac{1}{ad-bc}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}\). Substituting \(a=-6\), \(b=2\), \(c=8\), and \(d=-3\) into the formula, we find that the inverse of \(A\) is \(A^{-1}=\begin{bmatrix}-1.5 & -1\\-4 & -3\end{bmatrix}\).

Step 3 ：We can find the solution matrix \(X\) by multiplying the inverse of \(A\) with the constant matrix \(B\): \(X=A^{-1}B=\begin{bmatrix}-1.5 & -1\\-4 & -3\end{bmatrix}\begin{bmatrix}-10\\12\end{bmatrix}=\begin{bmatrix}3\\4\end{bmatrix}\).

Step 4 ：Thus, the solution to the system of equations is \(x=3\) and \(y=4\).

Step 5 ：Final Answer: The inverse of the coefficient matrix is \(A^{-1}=\boxed{\begin{bmatrix}-1.5 & -1\\-4 & -3\end{bmatrix}}\). The solution of the system is \(x=\boxed{3}\) and \(y=\boxed{4}\).