## Precalculus (6th Edition) Blitzer

The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 1 & -3 & 1 \\ 3 & -7 & 2 \\ \end{matrix} \right|\begin{matrix} 3 \\ 5 \\ 12 \\ \end{matrix} \right]$ Solve this matrix as below to get: $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 1 & -3 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} 3 \\ 5 \\ 3 \\ \end{matrix} \right]$ $By,\ 3{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 0 & -2 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} 3 \\ 7 \\ 3 \\ \end{matrix} \right]$ $By,\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 2 & -4 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & 0 \\ \end{matrix} \right|\begin{matrix} 3 \\ 7 \\ 4 \\ \end{matrix} \right]$ $By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ Convert the last row in equation form, \begin{align} 0x+0y+0z=4 & \\ 0=4 & \\ \end{align} And this statement is not true, so there is no solution of this system. Hence, there is no solution for this system of equations.