## Precalculus (6th Edition) Blitzer

Consider the following system of equations: \begin{align} & 2x-2y+2z=5 \\ & x-y+z=2 \\ & 2x+y-z=1 \end{align} The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 2 & -2 & 2 & 5 \\ 1 & -1 & 1 & 2 \\ 2 & 1 & -1 & 1 \\ \end{matrix} \right]$ Using the elementary row transformation we will obtain the echelon form of the matrix. ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & \frac{5}{2} \\ 1 & -1 & 1 & 2 \\ 2 & 1 & -1 & 1 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},\text{ }{{\text{R}}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & \frac{5}{2} \\ 0 & 0 & 0 & -\frac{1}{2} \\ 0 & 3 & -3 & -4 \\ \end{matrix} \right]$ ${{R}_{2}}\leftrightarrow {{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & \frac{5}{2} \\ 0 & 3 & -3 & -4 \\ 0 & 0 & 0 & -\frac{1}{2} \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+1{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 2 & \frac{5}{2} & 1 \\ 0 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Since, in the last row $0\ne 1$, this implies that the system is inconsistent, and has no solution.