## Precalculus (6th Edition) Blitzer

$\left\{ \left( 3+z,\ 5+5z,\ 3z+4,\ z \right) \right\}$
Convert the provided system of equations into matrix form. \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & 4 & 1 \\ \end{matrix} \\ & \begin{matrix} 1 & -1 & 3 & -5 \\ \end{matrix} \\ & \begin{matrix} 3 & 1 & -2 & -2 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 10 \\ 6 \\ \end{matrix} \right] Solve the above matrix as below: $\text{By},\ {{R}_{3}}-3{{R}_{2}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ 1 \\ \end{matrix} \\ & \begin{matrix} 1 & -1 & \ \ 3 & -5 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 4 & -11 & 13 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 10 \\ -24 \\ \end{matrix} \right] $\text{By},\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 1 & \ \ \ 2 & -11 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 4 & -11 & \ 13 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 13 \\ -24 \\ \end{matrix} \right] $By,\ {{R}_{3}}-4{{R}_{2}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 1 & \ \ \ 2 & -11 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 4 & -11 & \ 13 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 7 \\ 13 \\ -24 \\ \end{matrix} \right] Convert the above matrix system into the linear equations as below: \begin{align} & 2w-3x+4y+z=7 \\ & x+2y-11z=13 \\ & -19y+57z=-76 \end{align} Calculate the value of y in terms of z as below: \begin{align} & -19y+57z=-76 \\ & 19y=76+57z \\ & y=\frac{76}{19}+\frac{57}{19}z \\ & y=4+3z \end{align} Calculate the value of x in terms of z, by substituting the value ofy, z as below: \begin{align} & x+2\left( 4+3z \right)-11z=13 \\ & x+8+6z-11z=13 \\ & x=13-8+5z \\ & x=5+5z \end{align} Calculate the value of w in terms of z, by substituting the value of x, y, z as below: \begin{align} & 2w-3\left( 5+5z \right)+4\left( 4+3z \right)+z=7 \\ & 2w-15-15z+16+12z+z=7 \\ & w=\frac{6}{2}+\frac{2}{2}z \\ & w=3+z \end{align} The value of w, x and y in terms of z: \begin{align} & w=3+z \\ & x=5+5z \\ & y=3z+4 \\ & z=z \\ \end{align} Hence we have the solutions: $\left\{ \left( 3+z,\ 5+5z,\ 3z+4,\ z \right) \right\}$