Problem 4. Solving Systems of Equations with Inverses. 5 points. Write a matrix equation (i.e. $A X=B$ ) that is equivalent to the following system of linear equations: \[ \begin{aligned} -2 x-8 y & =16 \\ \frac{3}{4} x-3 y & =6 \end{aligned} \] then solve the system using the inverse (i.e. $X=A^{-1} B$ ). Solution:

Step 1 ：Convert the system of equations into a matrix equation. The coefficients of the variables x and y in the system of equations form the matrix A, the variables x and y form the matrix X, and the constants on the right side of the equations form the matrix B. So, we have \(A = \begin{bmatrix} -2 & -8 \\ 0.75 & -3 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \end{bmatrix}\), and \(B = \begin{bmatrix} 16 \\ 6 \end{bmatrix}\).

Step 2 ：Find the inverse of matrix A. The inverse of A is \(A^{-1} = \begin{bmatrix} -0.25 & 0.66666667 \\ -0.0625 & -0.16666667 \end{bmatrix}\).

Step 3 ：Solve for X by multiplying the inverse of A with B. This gives us \(X = A^{-1} B = \begin{bmatrix} -2.22044605e-16 \\ -2 \end{bmatrix}\).

Step 4 ：The first value is extremely close to zero, so we can consider it as zero for practical purposes. Therefore, the solution to the system of equations is \(x = 0\) and \(y = -2\).

Step 5 ：\(\boxed{X = \begin{bmatrix} 0 \\ -2 \end{bmatrix}}\)