Answer
No solution.
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&1&0\\ 3&2&-1&6\\ 1&4&-3&3 \end{bmatrix}
\begin{array} \\ \\R_2-3R_1\to R_2\\R_3-R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&1&0\\ 0&5&-4&6\\ 0&5&-4&3 \end{bmatrix}
\begin{array} \\ \\ \\-R_3+R_2\to R_3\\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&1&0\\ 0&5&-4&6\\ 0&0&0&3 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 4. Since the last row gives $0=3$, there is no solution to the system.
