Answer
$x=-20, y=10, z=16$
Work Step by Step
Step 1. Write the original system of equations in matrix form $AX=B$:
$\begin{bmatrix} 0 & -2 & 2\\3 & 1 & 3\\1 & -2 & 3 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}\begin{array}( \\= \\ \\ \end{array} \begin{bmatrix} 12\\-2\\8 \end{bmatrix}$
Step 2. Note that matrix A is the same as in Exercise 23. Recall the answer to that question, we have:
$\begin{array}(\\A^{-1}=\\ \\ \end{array} \begin{bmatrix} -9/2 & -1 & 4\\3 & 1 & -3\\7/2 & 1 & -3 \end{bmatrix} $
Step 3. Use the formula $X=A^{-1}B$, we get the solutions as:
$\begin{bmatrix} -9/2 & -1 & 4\\3 & 1 & -3\\7/2 & 1 & -3 \end{bmatrix} \begin{bmatrix} 12\\-2\\8 \end{bmatrix} \begin{array}(\\=\\ \\ \end{array} \begin{bmatrix} -20\\10\\16 \end{bmatrix}$
Step 4. Conclusion: the solutions for the system of equations are $x=-20, y=10, z=16$
