## Precalculus (6th Edition) Blitzer

The values of x, y, z are $6,-4,2$ respectively.
Let us assume a matrix $A$ be formed from the given system of equations, $A=\left[ \begin{matrix} 1 & -2 & 1 \\ 2 & -1 & -1 \\ 3 & 5 & -4 \\ \end{matrix} \right]$ Then, matrix $B$ is obtained by writing the right-hand side of the equations as follows: $B=\left[ \begin{matrix} 16 \\ 14 \\ -10 \\ \end{matrix} \right]$ Write the augmented matrix $\left[ A|B \right]$ as shown below: $\left[ \begin{matrix} 1 & -2 & 1 & 16 \\ 2 & -1 & -1 & 14 \\ 3 & 5 & -4 & -10 \\ \end{matrix} \right]$ Then, one can apply ${{R}_{2}}\to 2{{R}_{1}}-{{R}_{2}}$ and one gets: $\left[ \begin{matrix} 1 & -2 & 1 & 16 \\ 0 & -3 & 3 & 18 \\ 3 & 5 & -4 & -10 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to 3{{R}_{1}}-{{R}_{3}}$ and one gets: $\left[ \begin{matrix} 1 & -2 & 1 & 16 \\ 0 & -3 & 3 & 18 \\ 0 & -11 & 7 & 58 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to \frac{{{R}_{2}}}{-3}$ and one gets: $\left[ \begin{matrix} 1 & -2 & 1 & 16 \\ 0 & 1 & -1 & -6 \\ 0 & -11 & 7 & 58 \\ \end{matrix} \right]$ Apply ${{R}_{1}}\to {{R}_{1}}+2{{R}_{2}}$ and one gets: $\left[ \begin{matrix} 1 & 0 & -1 & 4 \\ 0 & 1 & -1 & -6 \\ 0 & -11 & 7 & 58 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to 11{{R}_{2}}+{{R}_{3}}$ and one gets: $\left[ \begin{matrix} 1 & 0 & -1 & 4 \\ 0 & 1 & -1 & -6 \\ 0 & 0 & -4 & -8 \\ \end{matrix} \right]$ Apply ${{R}_{1}}\to {{R}_{1}}-\frac{{{R}_{3}}}{4}$ and one gets: $\left[ \begin{matrix} 1 & 0 & 0 & 6 \\ 0 & 1 & -1 & -6 \\ 0 & 0 & -4 & -8 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to {{R}_{2}}-\frac{{{R}_{3}}}{4}$ and one gets: $\left[ \begin{matrix} 1 & 0 & 0 & 6 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & -4 & -8 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to \frac{{{R}_{3}}}{-4}$ and one gets: $\left[ \begin{matrix} 1 & 0 & 0 & 6 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 2 \\ \end{matrix} \right]$ Thus, the values of $x,y,z$ are $6,-4,2$.