## Precalculus (6th Edition) Blitzer

$\left\{ \left( \frac{-2z}{11}+\frac{81}{11},\ \frac{z}{22}+\frac{10}{11},\ \frac{4z}{11}-\frac{8}{11},\ z \right) \right\}$
Convert the provided system of equations into matrix form. \left[ \left. \begin{align} & \begin{matrix} 1 & \ 2 & \ \ 3 & -1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 2 & -3 & \ 1 \\ \end{matrix} \\ & \begin{matrix} 1 & -4 & \ 1 & \ 0 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 4 \\ 3 \\ \end{matrix} \right] Solve the above matrix as below: $\text{By},\ {{R}_{1}}-{{R}_{3}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 1 & \ 2 & \ \ 3 & \ -1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 2 & -3 & \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 6 & \ \ 2 & \ -1 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 4 \\ 4 \\ \end{matrix} \right] $\text{By},\ {{R}_{3}}-3{{R}_{2}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 1 & \ 2 & \ \ 3 & \ \ -1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 2 & -3 & \ \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 0 & \ \ 11 & \ -4 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 4 \\ -8 \\ \end{matrix} \right] Convert the above matrix system into the linear equations. \begin{align} w+2x+3y-z=7 & \\ 2x-3y+z=4 & \\ 11y-4z=-8 & \\ \end{align} Calculate the value of y in terms of z as below: \begin{align} & 11y-4z=-8 \\ & 11y=-8+4z \\ & y=\frac{-8}{11}+\frac{4}{11}z \end{align} Calculate the value of x in terms of z, by substituting the value of y, z as below: \begin{align} & 2x-3\left( \frac{-8}{11}+\frac{4}{11}z \right)+z=4 \\ & 2x+\frac{24}{11}-\frac{12}{11}z+z=4 \\ & 2x=4-\frac{24}{11}+\frac{1}{11}z \\ & x=\frac{10}{11}+\frac{1}{22}z \end{align} Calculate the value of w in terms of z, by substituting the value of x, y, z as below: \begin{align} & w+2\left( \frac{10}{11}+\frac{1}{22}z \right)+3\left( \frac{-8}{11}+\frac{4}{11}z \right)-z=7 \\ & w+\frac{20}{11}+\frac{2z}{22}-\frac{24}{11}+\frac{12z}{11}-z=7 \\ & w=7+\frac{4}{11}-\frac{2}{11}z \\ & w=\frac{81}{11}-\frac{2}{11}z \end{align} The value of w, x and y in terms of z are, \begin{align} & w=\frac{-2z}{11}+\frac{81}{11} \\ & x=\frac{10}{11}+\frac{z}{22} \\ & y=\frac{4z}{11}-\frac{8}{11} \\ & z=z \\ \end{align} Hence the numbers satisfying the set of solutions are: $\left\{ \left( \frac{-2z}{11}+\frac{81}{11},\ \frac{z}{22}+\frac{10}{11},\ \frac{4z}{11}-\frac{8}{11},\ z \right) \right\}$