## Precalculus (6th Edition) Blitzer

The values of w, x, y, z are $w=1,x=3,y=0,z=-2$.
Consider the system of the equations: \begin{align} & 2w+y-3z=8 \\ & w-x+4z=-10 \\ & 3w+5x-y-z=20 \\ & w+x-y-z=6 \end{align} The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 2 & 0 & 1 & -3 & 8 \\ 1 & -1 & 0 & 4 & -10 \\ 3 & 5 & -1 & -1 & 20 \\ 1 & 1 & -1 & -1 & 6 \\ \end{matrix} \right]$ The matrix will be converted into the echelon form by using the elementary row transformation. ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 1 & -1 & 0 & 4 & -10 \\ 3 & 5 & -1 & -1 & 20 \\ 1 & 1 & -1 & -1 & 6 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -3 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -1 \right){{R}_{1}}$ give, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & -1 & -\frac{1}{2} & \frac{11}{2} & -14 \\ 0 & 5 & -\frac{5}{2} & \frac{7}{2} & 8 \\ 0 & 1 & -\frac{3}{2} & \frac{1}{2} & 2 \\ \end{matrix} \right]$ ${{R}_{2}}\to -1{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\ 0 & 5 & -\frac{5}{2} & \frac{7}{2} & 8 \\ 0 & 1 & -\frac{3}{2} & \frac{1}{2} & 2 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -5 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -1 \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\ 0 & 0 & -5 & 31 & -62 \\ 0 & 0 & -2 & 6 & -12 \\ \end{matrix} \right]$ ${{R}_{3}}\to -\frac{1}{5}{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\ 0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\ 0 & 0 & -2 & 6 & -12 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+\left( 2 \right){{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\ 0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\ 0 & 0 & 0 & -\frac{32}{5} & \frac{64}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to -\frac{5}{32}{{R}_{4}}$ gives, $\left[ \begin{matrix} 1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\ 0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\ 0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\ 0 & 0 & 0 & 1 & -2 \\ \end{matrix} \right]$ As, the desired matrix is in row echelon form, so, the system of the linear equations corresponding to the echelon form of the matrix will be as below: $w+\frac{1}{2}y-\frac{3}{2}z=4$ (I) $x+\frac{1}{2}y-\frac{11}{2}z=14$ (II) $y-\frac{31}{5}z=\frac{62}{5}$ (III) $z=-2$ (IV) Apply the back-substitution method: Equation (IV) gives: $z=-2$. Put the value of z in the equation (III) as follows: \begin{align} & y-\frac{31}{5}\left( -2 \right)=\frac{62}{5} \\ & y=\frac{62}{5}-\frac{62}{5} \\ & y=0 \end{align} Put the values of y and z in the equation (II) as follows: \begin{align} & x+\frac{1}{2}\left( 0 \right)-\frac{11}{2}\left( -2 \right)=14 \\ & x=14-11 \\ & x=3 \end{align} Put the values of x, y, and z in the equation (I) as follows: \begin{align} & w+\frac{1}{2}\left( 0 \right)-\frac{3}{2}\left( -2 \right)=4 \\ & w=4-3 \\ & w=1 \end{align} Therefore, the values of w, x, y, z are $w=1,x=3,y=0,z=-2$.