Answer
$(0,1,2)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&2&6\\ 1&-1&0&-1\\ 2&1&3&7 \end{bmatrix}
\begin{array} \\ \\-R_2+R_1\to R_2\\-R_3+2R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&2&6\\ 0&3&2&7\\ 0&3&1&5 \end{bmatrix}
\begin{array} \\ \\ \\-R_3+R_2\to R_3\\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&2&6\\ 0&3&2&7\\ 0&0&1&2 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 4. The last row gives $z=2$. Back substitute into the second row to get $3y+2\times2=7$ or $y=1$.
Step 5. Continue the back substitution process to get the solution as $(0,1,2)$
