## Precalculus (6th Edition) Blitzer

The system solution set is $\left( 1,3,2,1 \right)$.
Consider the given system of the equation: \left\{ \begin{align} & 2w+x-y=3 \\ & w-3x+2y=-4 \\ & 3w+x-3y+z=1 \\ & w+2x-4y-z=-2 \end{align} \right. The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 2 & 1 & -1 & 0 & 3 \\ 1 & -3 & 2 & 0 & -4 \\ 3 & 1 & -3 & 1 & 1 \\ 1 & 2 & -4 & -1 & -2 \\ \end{matrix} \right]$ Using the elementary row transformation we will find the echelon form of the matrix. Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 2 & 1 & -1 & 0 & 3 \\ 3 & 1 & -3 & 1 & 1 \\ 1 & 2 & -4 & -1 & -2 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 10 & -9 & 1 & 13 \\ 1 & 2 & -4 & -1 & -2 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}-{{R}_{1}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 10 & -9 & 1 & 13 \\ 0 & 5 & -6 & -1 & 2 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-2{{R}_{4}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 0 & 3 & 3 & 9 \\ 0 & 5 & -6 & -1 & 2 \\ \end{matrix} \right]$ Interchange ${{R}_{3}}\leftrightarrow {{R}_{4}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 5 & -6 & -1 & 2 \\ 0 & 0 & 3 & 3 & 9 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-\frac{5}{7}{{R}_{2}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 0 & -\frac{17}{7} & -1 & -\frac{41}{7} \\ 0 & 0 & 3 & 3 & 9 \\ \end{matrix} \right]$ Apply ${{R}_{4}}\to {{R}_{4}}+\frac{21}{17}{{R}_{3}}$ $\left[ \begin{matrix} 1 & -3 & 2 & 0 & -4 \\ 0 & 7 & -5 & 0 & 11 \\ 0 & 0 & -\frac{17}{7} & -1 & -\frac{41}{7} \\ 0 & 0 & 0 & \frac{30}{17} & \frac{30}{17} \\ \end{matrix} \right]$ Now use back-substitution to express $w,x,y,z$. \begin{align} & w-3x+2y=-4 \\ & 7x-5y=11 \\ & -\frac{17}{7}y-z=-\frac{41}{7} \\ & z=1 \end{align} Substitute $z=1$ in the third equation to get: \begin{align} & -\frac{17}{7}y-1=-\frac{41}{7} \\ & -\frac{17}{7}y=-\frac{41}{7}+1 \\ & -\frac{17}{7}y=-\frac{34}{7} \\ & y=2 \end{align} Substitute $y=2$ in the second equation to get: \begin{align} & 7x-5\left( 2 \right)=11 \\ & 7x=11+10 \\ & x=3 \end{align} Substitute $x=3$ and $y=2$ in the first equation to get: \begin{align} & w-3\left( 3 \right)+2\left( 2 \right)=-4 \\ & w-9+4=-4 \\ & w=-4+5 \\ & w=1 \end{align} Hence, the system’s solution set is $\left( 1,3,2,1 \right)$.