Answer
No solution.
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&3&2\\ 2&-1&-5&1\\ 4&3&1&6 \end{bmatrix}
\begin{array} \\ \\-R_2+2R_1\to R_2\\-R_3+4R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&3&2\\ 0&5&11&3\\ 0&5&11&2 \end{bmatrix}
\begin{array} \\ \\ \\-R_3+R_2\to R_3\\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&2&3&2\\ 0&5&11&3\\ 0&0&0&1 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 4. Since the last row gives $0=1$, there is no solution to the system.
