## Precalculus (6th Edition) Blitzer

The values of w, x, y, z are $w=1,x=2,y=3,z=-2$.
Consider the system of equations: \begin{align} & w+x+y+z=4 \\ & 2w+x-2y-z=0 \\ & w-2x-y-2z=-2 \\ & 3w+2x+y+3z=4 \end{align} The matrix corresponding to the system of equation is as follows: $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 2 & 1 & -2 & -1 & 0 \\ 1 & -2 & -1 & -2 & -2 \\ 3 & 2 & 1 & 3 & 4 \\ \end{matrix} \right]$ The matrix will be converted into the echelon form by using the elementary row transformation. ${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -3 \right){{R}_{1}}$ give, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & -1 & -4 & -3 & -8 \\ 0 & -3 & -2 & -3 & -6 \\ 0 & -1 & -2 & 0 & -8 \\ \end{matrix} \right]$ ${{R}_{2}}\to \left( -1 \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 4 & 3 & 8 \\ 0 & -3 & -2 & -3 & -6 \\ 0 & -1 & -2 & 0 & -8 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( 3 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( 1 \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 4 & 3 & 8 \\ 0 & 0 & 10 & 6 & 18 \\ 0 & 0 & 2 & 3 & 0 \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{1}{10}{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 4 & 3 & 8 \\ 0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\ 0 & 0 & 2 & 3 & 0 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+\left( -2 \right){{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 4 & 3 & 8 \\ 0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\ 0 & 0 & 0 & \frac{9}{5} & -\frac{18}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{5}{9}{{R}_{4}}$ gives, $\left[ \begin{matrix} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 4 & 3 & 8 \\ 0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\ 0 & 0 & 0 & 1 & -2 \\ \end{matrix} \right]$ Since the desired matrix is now in row echelon form, so, express the system of linear equations corresponding to the echelon form of the matrix as follows: $w+x+y+z=4$ (I) $x+4y+3z=8$ (II) $y+\frac{3}{5}z=\frac{9}{5}$ (III) $z=-2$ (IV) Apply the back-substitution method: Equation (IV) gives, $z=-2$. Substitute the value of z in the equation (III) as follows: \begin{align} & y+\frac{3}{5}\left( -2 \right)=\frac{9}{5} \\ & y=\frac{9}{5}+\frac{6}{5} \\ & y=3 \end{align} Substitute the values of y and z in the equation (II) as follows: \begin{align} & x+4\left( 3 \right)+3\left( -2 \right)=8 \\ & x=2 \end{align} Substitute the values of x, y, and z in the equation (I) as follows: \begin{align} & w+\left( 2 \right)+\left( 3 \right)+\left( -2 \right)=4 \\ & w=1 \end{align} Therefore, the values of w, x, y, z are $w=1,x=2,y=3,z=-2$.