## Precalculus (6th Edition) Blitzer

$\left\{ \left( 2,\ -3,\ 7 \right) \right\}$
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 1 & 2 & 1 \\ 3 & 4 & 2 \\ \end{matrix} \right|\begin{matrix} 0 \\ 3 \\ 8 \\ \end{matrix} \right]$ Now we will solve this matrix as below to get: $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ 3 \\ 1 \\ \end{matrix} \right]$ $By,\ 3{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 0 & 5 & 3 \\ 0 & 2 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ 6 \\ 1 \\ \end{matrix} \right]$ $By,\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 2 & -1 & -1 \\ 0 & 5 & 3 \\ 0 & 1 & 0 \\ \end{matrix} \right|\begin{matrix} 0 \\ 6 \\ -3 \\ \end{matrix} \right]$ $By,\ 3{{R}_{3}}-{{R}_{2}}\to {{R}_{3}}$ Convert the matrix into equation form: \begin{align} 2x-y-z=0 & \\ 5y+3z=6 & \\ y=-3 & \\ \end{align} Substitute the value of $y$ in $5y+3z=6$ to obtain the value of $z$: \begin{align} & 5\times \left( -3 \right)+3z=6 \\ & 3z=6+15 \\ & 3z=21 \\ & z=7 \end{align} Substitute the value of $y,z$ in $2x-y-z=0$ to obtain the value of $x$: \begin{align} & 2x-\left( -3 \right)-7=0 \\ & 2x=-3+7 \\ & 2x=4 \\ & x=2 \end{align} Hence, the solution is $\left\{ \left( 2,\ -3,\ 7 \right) \right\}$