## Precalculus (6th Edition) Blitzer

The solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$.
Consider the given system of equations: \left\{ \begin{align} & 2w-x+3y+z=0 \\ & 3w+2x+4y-z=0 \\ & 5w-2x-2y-z=0 \\ & 2w+3x-7y-5z=0 \end{align} \right. The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 2 & -1 & 3 & 1 & 0 \\ 3 & 2 & 4 & -1 & 0 \\ 5 & -2 & -2 & -1 & 0 \\ 2 & 3 & -7 & -5 & 0 \\ \end{matrix} \right]$ Use the elementary row transformation to find the echelon form of the matrix. Apply ${{R}_{2}}\to {{R}_{2}}-{{R}_{4}}$ $\left[ \begin{matrix} 2 & -1 & 3 & 1 & 0 \\ 1 & -1 & 11 & 4 & 0 \\ 5 & -2 & -2 & -1 & 0 \\ 2 & 3 & -7 & -5 & 0 \\ \end{matrix} \right]$ Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$ $\left[ \begin{matrix} 1 & -1 & 11 & 4 & 0 \\ 2 & -1 & 3 & 1 & 0 \\ 5 & -2 & -2 & -1 & 0 \\ 2 & 3 & -7 & -5 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}-{{R}_{2}}$, ${{R}_{3}}\to {{R}_{3}}-5{{R}_{1}}$ $\left[ \begin{matrix} 1 & -1 & 11 & 4 & 0 \\ 2 & -1 & 3 & 1 & 0 \\ 0 & 3 & -57 & -21 & 0 \\ 0 & 4 & -10 & -6 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ $\left[ \begin{matrix} 1 & -1 & 11 & 4 & 0 \\ 0 & 1 & -19 & -7 & 0 \\ 0 & 3 & -57 & -21 & 0 \\ 0 & 4 & -10 & -6 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{2}},{{R}_{4}}\to {{R}_{4}}-4{{R}_{2}}$ $\left[ \begin{matrix} 1 & -1 & 11 & 4 & 0 \\ 0 & 1 & -19 & -7 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 66 & 22 & 0 \\ \end{matrix} \right]$ In third row, after back substitution, $0w+0x+0y+0z=0$ This means the system has infinitely many solutions. Express $w,x,y\text{ and }z$ in terms of z: From fourth row: \begin{align} & 66y+22z=0 \\ & 3y+z=0 \\ & y=-\frac{1}{3}z \end{align} From second row we get; \begin{align} & x-19y-7z=0 \\ & x-19\left( -\frac{1}{3}z \right)-7z=0 \\ & x-\frac{2}{3}z=0 \\ & x=\frac{2}{3}z \end{align} From first row we get: \begin{align} & w-x+11y+4z=0 \\ & w-\frac{2}{3}z+11\left( -\frac{1}{3}z \right)+4z=0 \\ & w-\frac{1}{3}z=0 \\ & w=\frac{1}{3}z \end{align} Let $z=\text{ any variable say, }t$; the solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$. Thus, the solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$.