Answer
$(10, -10, -5)$
Work Step by Step
Step 1. Define the following matrices:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 2&0&3\\ 1&1&6\\3&-1&1\end{bmatrix},
\begin{array} \\X=\\ \end{array}
\begin{bmatrix} x\\ y\\z \end{bmatrix},
\begin{array} \\B=\\ \end{array}
\begin{bmatrix} 5\\0\\5\end{bmatrix}$
Step 2. Rewrite the original system of equations as
$AX=B$, thus $A^{-1}AX=X=A^{-1}B$
Step 3. Find the inverse $A^{-1}$ using the procedure given in section 10.5:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&0&3&|&1&0&0\\ 1&1&6&|&0&1&0\\3&-1&1&|&0&0&1\end{bmatrix}
\begin{array} \\ \\2R_2-R_1\to R_2\\ 3R_2-R_3\to R_3 \end{array}$
Step 4. Perform the row operations (row3 first):
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&0&3&|&1&0&0\\ 0&2&9&|&-1&2&0\\0&4&17&|&0&3&-1\end{bmatrix}
\begin{array} \\ \\ \\ 2R_2-R_3\to R_3 \end{array}$
Step 5. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&0&3&|&1&0&0\\ 0&1&0&|&-2&1&0\\0&0&1&|&-2&1&1\end{bmatrix}
\begin{array} ((R_1-3R_3)/2\to R_1\\ \\ \\ \end{array}$
Step 6. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&|&7/2&-3/2&-3/2\\ 0&1&0&|&-2&1&0\\0&0&1&|&-2&1&1\end{bmatrix}$
Step 7. We obtain the inverse:
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 7/2&-3/2&-3/2\\ -2&1&0\\-2&1&1\end{bmatrix}$
Step 8. The solutions:
$\begin{array} \\X=A^{-1}B=\\ \end{array}
\begin{bmatrix} 7/2&-3/2&-3/2\\ -2&1&0\\-2&1&1\end{bmatrix}
\begin{bmatrix} 5\\0\\5\end{bmatrix}
=\begin{bmatrix} 10\\-10\\-5\end{bmatrix}$
or $(10, -10, -5)$
