## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 893: 37

#### Answer

The values of w, x, y, z are $w=0,x=-3,y=0,z=-3$.

#### Work Step by Step

Consider the system of the equations: \begin{align} & 3w-4x+y+z=9 \\ & w+x-y-z=0 \\ & 2w+x+4y-2z=3 \\ & -w+2x+y-3z=3 \end{align} The matrix corresponding to the system of equation is as follows: $\left[ \begin{matrix} 3 & -4 & 1 & 1 & 9 \\ 1 & 1 & -1 & -1 & 0 \\ 2 & 1 & 4 & -2 & 3 \\ -1 & 2 & 1 & -3 & 3 \\ \end{matrix} \right]$ The matrix will be converted into the echelon form by using the elementary row transformation. ${{R}_{1}}\to \frac{1}{3}{{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 1 & 1 & -1 & -1 & 0 \\ 2 & 1 & 4 & -2 & 3 \\ -1 & 2 & 1 & -3 & 3 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( 1 \right){{R}_{1}}$ give, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & \frac{7}{3} & -\frac{4}{3} & -\frac{4}{3} & -3 \\ 0 & \frac{11}{3} & \frac{10}{3} & -\frac{8}{3} & -3 \\ 0 & \frac{2}{3} & \frac{4}{3} & -\frac{8}{3} & 6 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{3}{7}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\ 0 & \frac{11}{3} & \frac{10}{3} & -\frac{8}{3} & -3 \\ 0 & \frac{2}{3} & \frac{4}{3} & -\frac{8}{3} & 6 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -\frac{11}{3} \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -\frac{2}{3} \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\ 0 & 0 & \frac{38}{7} & -\frac{4}{7} & \frac{12}{7} \\ 0 & 0 & \frac{12}{7} & -\frac{16}{7} & \frac{48}{7} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{7}{38}{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\ 0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\ 0 & 0 & \frac{12}{7} & -\frac{16}{7} & \frac{48}{7} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+\left( -\frac{12}{7} \right){{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\ 0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\ 0 & 0 & 0 & -\frac{40}{19} & \frac{120}{19} \\ \end{matrix} \right]$ ${{R}_{4}}\to -\frac{19}{40}{{R}_{4}}$ gives, $\left[ \begin{matrix} 1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\ 0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\ 0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ Since the desired matrix is now in row echelon form, so express, the system of linear equations corresponding to the echelon form of matrix as follows: $w-\frac{4}{3}x+\frac{1}{3}y+\frac{1}{3}z=3$ (I) $x-\frac{4}{7}y-\frac{4}{7}z=-\frac{9}{7}$ (II) $y-\frac{2}{19}z=\frac{6}{19}$ (III) $z=-3$ (IV) Apply the back-substitution method: Equation (4) gives, $z=-3$. Substitute the value of z in the equation (3) as follows: \begin{align} & y-\frac{2}{19}\left( -3 \right)=\frac{6}{19} \\ & y=0 \end{align} Substitute the values of y and z in the equation (2)as follows: \begin{align} & x-\frac{4}{7}\left( 0 \right)-\frac{4}{7}\left( -3 \right)=-\frac{9}{7} \\ & x=-3 \end{align} Substitute the values of x, y, and z in the equation (1)as follows: \begin{align} & w-\frac{4}{3}\left( -3 \right)+\frac{1}{3}\left( 0 \right)+\frac{1}{3}\left( -3 \right)=3 \\ & w=0 \end{align} Therefore, the values of w, x, y, z are $w=0,x=-3,y=0,z=-3$.

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