## Precalculus (6th Edition) Blitzer

The solution is, $w=1,x=-2,y=3\text{ and }z=-4$.
Write the augmented matrix, $\left[ \begin{matrix} 3 & 2 & -1 & 2 \\ 4 & -1 & 1 & 2 \\ 1 & 1 & 1 & 1 \\ -2 & 3 & 2 & -3 \\ \end{matrix}\left| \begin{matrix} -12 \\ 1 \\ -2 \\ 10 \\ \end{matrix} \right. \right]$ \begin{align} & {{R}_{1}}\leftrightarrow {{R}_{3}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 4 & -1 & 1 & 2 \\ 3 & 2 & -1 & 2 \\ -2 & 3 & 2 & -3 \\ \end{matrix}\left| \begin{matrix} -2 \\ 1 \\ -12 \\ 10 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{2}}\to {{R}_{2}}-4{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}+{{R}_{3}}\text{,}{{R}_{4}}\to {{R}_{4}}\text{+}2{{R}_{1}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & -5 & -3 & -2 \\ 0 & -1 & -4 & -1 \\ 0 & 5 & 4 & -1 \\ \end{matrix}\left| \begin{matrix} -2 \\ 9 \\ -6 \\ 6 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{2}}\to -\frac{1}{5}{{R}_{2}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{5} & \frac{2}{5} \\ 0 & -1 & -4 & -1 \\ 0 & 5 & 4 & -1 \\ \end{matrix}\left| \begin{matrix} -2 \\ -\frac{9}{5} \\ -6 \\ 6 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{3}}\to {{R}_{3}}+{{R}_{2}},{{R}_{4}}\to {{R}_{4}}-5{{R}_{2}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{5} & \frac{2}{5} \\ 0 & 0 & -\frac{17}{5} & -\frac{3}{5} \\ 0 & 0 & 1 & -3 \\ \end{matrix}\left| \begin{matrix} -2 \\ -\frac{9}{5} \\ -\frac{39}{5} \\ 15 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{3}}\to -\frac{5}{17}{{R}_{3}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{5} & \frac{2}{5} \\ 0 & 0 & 1 & \frac{3}{17} \\ 0 & 0 & 1 & -3 \\ \end{matrix}\left| \begin{matrix} -2 \\ -\frac{9}{5} \\ \frac{39}{17} \\ 15 \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{4}}\to {{R}_{4}}-{{R}_{3}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{5} & \frac{2}{5} \\ 0 & 0 & 1 & \frac{3}{17} \\ 0 & 0 & 0 & -\frac{54}{17} \\ \end{matrix}\left| \begin{matrix} -2 \\ -\frac{9}{5} \\ \frac{39}{17} \\ \frac{216}{17} \\ \end{matrix} \right. \right] \\ \end{align} \begin{align} & {{R}_{4}}\to -\frac{17}{54}{{R}_{4}} \\ & \left[ \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{5} & \frac{2}{5} \\ 0 & 0 & 1 & \frac{3}{17} \\ 0 & 0 & 0 & 1 \\ \end{matrix}\left| \begin{matrix} -2 \\ -\frac{9}{5} \\ \frac{39}{17} \\ -4 \\ \end{matrix} \right. \right] \\ \end{align} Therefore, the original system is equivalent to, \begin{align} & w+x+y+z=-2 \\ & x+\frac{3}{5}y+\frac{2}{5}z=-\frac{9}{5} \\ & y+\frac{3}{17}z=\frac{39}{17} \\ & z=-4 \\ \end{align} From the last equation: $z=-4$ Substitute zin $y+\frac{3}{17}z=\frac{39}{17}$: \begin{align} & y-\frac{12}{17}=\frac{39}{17} \\ & y=\frac{51}{17} \\ & y=3 \\ \end{align} Substitute z, y in $x+\frac{3}{5}y+\frac{2}{5}z=-\frac{9}{5}$ \begin{align} & x+\frac{9}{5}-\frac{8}{5}=-\frac{9}{5} \\ & x+\frac{1}{5}=-\frac{9}{5} \\ & x=-\frac{10}{5} \\ & x=-2 \\ \end{align} Substitute x, y, z in $w+x+y+z=-2$ \begin{align} & w-2+3-4=-2 \\ & w+3=4 \\ & w=1 \\ \end{align} The value of variables is: \begin{align} & w=1 \\ & x=-2 \\ & y=3 \\ & z=-4 \end{align} Hence, the solution is $w=1,x=-2,y=3\text{ and }z=-4$.