## Precalculus (6th Edition) Blitzer

The system solution set is $\left( -1,2,1,1 \right)$.
Consider the given system of equations, \left\{ \begin{align} & w-2x-y-3z=-9 \\ & w+x-y=0 \\ & 3w+4x+z=6 \\ & 2x-2y+z=3 \end{align} \right. The given system of equations can be written in matrix form as below: $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 1 & 1 & -1 & 0 & 0 \\ 3 & 4 & 0 & 1 & 6 \\ 0 & 2 & -2 & 1 & 3 \\ \end{matrix} \right]$ Using the elementary row transformation we will find the echelon form of the matrix. Apply ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$: $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 0 & 3 & 0 & 3 & 9 \\ 0 & 10 & 3 & 10 & 33 \\ 0 & 2 & -2 & 1 & 3 \\ \end{matrix} \right]$ Now, apply ${{R}_{3}}\to {{R}_{3}}-5{{R}_{4}}$: $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 0 & 3 & 0 & 3 & 9 \\ 0 & 0 & 13 & 5 & 18 \\ 0 & 2 & -2 & 1 & 3 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to \frac{1}{3}{{R}_{2}}:$ $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 0 & 1 & 0 & 1 & 3 \\ 0 & 0 & 13 & 5 & 18 \\ 0 & 2 & -2 & 1 & 3 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}-2{{R}_{2}}:$ $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 0 & 1 & 0 & 1 & 3 \\ 0 & 0 & 13 & 5 & 18 \\ 0 & 0 & -2 & -1 & -3 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}+\frac{2}{13}{{R}_{3}}:$ $\left[ \begin{matrix} 1 & -2 & -1 & -3 & -9 \\ 0 & 1 & 0 & 1 & 3 \\ 0 & 0 & 13 & 5 & 18 \\ 0 & 0 & 0 & -\frac{3}{13} & -\frac{3}{13} \\ \end{matrix} \right]$ This gives the system as: \begin{align} & w-2x-y-3z=-9 \\ & x+z=3 \\ & 13y+5z=18 \\ & z=1 \end{align} Substitute $z=1$ in the third equation to get: \begin{align} & 13y+5\left( 1 \right)=18 \\ & 13y=18-5 \\ & 13y=13 \\ & y=1 \end{align} Substitute $z=1$ in the second equation to get: \begin{align} & x+1=3 \\ & x=2 \end{align} Substitute $z=1$, $x=2$, and $y=1$ in the first equation to get: \begin{align} & w-2\left( 2 \right)-1-3\left( 1 \right)=-9 \\ & w-4-4=-9 \\ & w=-1 \end{align} The system’s solution set is: $\left( -1,2,1,1 \right)$.