## Precalculus (6th Edition) Blitzer

The system solution set is $\left( 1,-2,1,1 \right)$.
Consider the given system of equations: \left\{ \begin{align} & w-3x+y-4z=4 \\ & -2w+x+2y=-2 \\ & 3w-2x+y-6z=2 \\ & -w+3x+2y-z=-6 \end{align} \right. The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 1 & -3 & 1 & -4 & 4 \\ -2 & 1 & 2 & 0 & -2 \\ 3 & -2 & 1 & -6 & 2 \\ -1 & 3 & 2 & -1 & -6 \\ \end{matrix} \right]$ Use the elementary row transformation to find the echelon form of the matrix. Apply ${{R}_{2}}\to {{R}_{2}}+2{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ $\left[ \begin{matrix} 1 & -3 & 1 & -4 & 4 \\ 0 & -5 & 4 & -8 & 6 \\ 0 & 7 & -2 & 6 & -10 \\ -1 & 3 & 2 & -1 & -6 \\ \end{matrix} \right]$ Now, apply ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}$ $\left[ \begin{matrix} 1 & -3 & 1 & -4 & 4 \\ 0 & -5 & 4 & -8 & 6 \\ 0 & 7 & -2 & 6 & -10 \\ 0 & 0 & 3 & -5 & -2 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}+\frac{7}{5}{{R}_{2}}$ $\left[ \begin{matrix} 1 & -3 & 1 & -4 & 4 \\ 0 & -5 & 4 & -8 & 6 \\ 0 & 0 & \frac{18}{5} & -\frac{26}{5} & -\frac{8}{5} \\ 0 & 0 & 3 & -5 & -2 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}-\frac{15}{18}{{R}_{3}}$ $\left[ \begin{matrix} 1 & -3 & 1 & -4 & 4 \\ 0 & -5 & 4 & -8 & 6 \\ 0 & 0 & \frac{18}{5} & -\frac{26}{5} & -\frac{8}{5} \\ 0 & 0 & 0 & -\frac{2}{3} & -\frac{2}{3} \\ \end{matrix} \right]$ Now, use back-substitution to express $w,x,y,z$. \begin{align} & w-3x+1y-4z=4 \\ & -5x+4y-8z=6 \\ & \frac{18}{5}y-\frac{26}{5}z=-\frac{8}{5} \\ & z=1 \end{align} Substitute $z=1$ in the third equation as follows; \begin{align} & \frac{18}{5}y-\frac{26}{5}\left( 1 \right)=-\frac{8}{5} \\ & \frac{18}{5}y=-\frac{8}{5}+\frac{26}{5} \\ & \frac{18}{5}y=\frac{18}{5} \\ & y=1 \end{align} Substitute $y=1$, $z=1$ in the second equation as follows: \begin{align} & -5x+4\left( 1 \right)-8\left( 1 \right)=6 \\ & -5x=6+4 \\ & x=-2 \end{align} Substitute $x=-2$, $y=1$, $z=1$ in the first equation as follows: \begin{align} & w-3\left( -2 \right)+1\left( 1 \right)-4\left( 1 \right)=4 \\ & w+6-3=4 \\ & w=4-3 \\ & w=1 \end{align} Hence, the system’s solution set is $\left( 1,-2,1,1 \right)$.