## Precalculus (6th Edition) Blitzer

The system solution set is $\phi$.
Consider the system of equations: \left\{ \begin{align} & 2w+x-2y-z=3 \\ & w-2x+y+z=4 \\ & -w-8x+7y+5z=13 \\ & 3w+x-2y+2z=6 \end{align} \right. The given system of equations can be written in matrix form as below: $\left[ \begin{matrix} 2 & 1 & -2 & -1 & 3 \\ 1 & -2 & 1 & 1 & 4 \\ -1 & -8 & 7 & 5 & 13 \\ 3 & 1 & -2 & 2 & 6 \\ \end{matrix} \right]$ Using the elementary row transformation we will find the echelon form of the matrix. Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$: $\left[ \begin{matrix} 1 & -2 & 1 & 1 & 4 \\ 2 & 1 & -2 & -1 & 3 \\ -1 & -8 & 7 & 5 & 13 \\ 3 & 1 & -2 & 2 & 6 \\ \end{matrix} \right]$ Now, apply ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}$, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ $\left[ \begin{matrix} 1 & -2 & 1 & 1 & 4 \\ 0 & 5 & -4 & -3 & -5 \\ 0 & -10 & 8 & 6 & 17 \\ 3 & 1 & -2 & 2 & 6 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}+2{{R}_{2}}$ $\left[ \begin{matrix} 1 & -2 & 1 & 1 & 4 \\ 0 & 5 & -4 & -3 & -5 \\ 0 & 0 & 0 & 0 & 7 \\ 3 & 1 & -2 & 2 & 6 \\ \end{matrix} \right]$ In third row, after back substitution, $0w+0x+0y+0z=7$ Which is not possible; therefore the linear system has no solution. Hence, the system solution set is $\phi$.