Answer
$(-\frac{1}{12},\frac{1}{12}, \frac{1}{12})$
Work Step by Step
Step 1. Define the following matrices:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 2&1&5\\ 1&2&2\\1&0&3\end{bmatrix},
\begin{array} \\X=\\ \end{array}
\begin{bmatrix} x\\ y\\z \end{bmatrix},
\begin{array} \\B=\\ \end{array}
\begin{bmatrix} 1/3\\1/4\\1/6\end{bmatrix}$
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:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&1&5&|&1&0&0\\ 1&2&2&|&0&1&0\\1&0&3&|&0&0&1\end{bmatrix}
\begin{array} \\ \\R_2-R_3\to R_2\\ R_1-2R_3\to R_3 \end{array}$
Step 4. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&1&5&|&1&0&0\\ 0&2&-1&|&0&1&-1\\0&1&-1&|&1&0&-2\end{bmatrix}
\begin{array} \\ \\ \\ -2R_3+R_2\to R_3 \end{array}$
Step 5. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&1&5&|&1&0&0\\ 0&2&-1&|&0&1&-1\\0&0&1&|&-2&1&3\end{bmatrix}
\begin{array} ( R_1-5R_3\to R_1\\ (R_3+R_2)/2\to R_3 \\ \\ \end{array}$
Step 6. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 2&1&0&|&11&-5&-15\\ 0&1&0&|&-1&1&1\\0&0&1&|&-2&1&3\end{bmatrix}
\begin{array} ( (R_1-R_2)/2\to R_1\\ \\ \\ \end{array}$
Step 7. Perform the row operations:
$\begin{array} \\AI=\\ \end{array}
\begin{bmatrix} 1&0&0&|&13/2&-3&-9\\ 0&1&0&|&-1&1&1\\0&0&1&|&-2&1&3\end{bmatrix}$
Step 8. We obtain the inverse:
$\begin{array} \\A^{-1}=\\ \end{array}
\begin{bmatrix} 13/2&-3&-9\\ -1&1&1\\-2&1&3\end{bmatrix}$
Step 9. The solutions:
$\begin{array} \\X=A^{-1}B=\\ \end{array}
\begin{bmatrix} 13/2&-3&-9\\ -1&1&1\\-2&1&3\end{bmatrix}
\begin{bmatrix} 1/3\\1/4\\1/6\end{bmatrix}
=\begin{bmatrix} -\frac{1}{12}\\\frac{1}{12}\\\frac{1}{12}\end{bmatrix}$
or $(-\frac{1}{12},\frac{1}{12}, \frac{1}{12})$
