Precalculus (6th Edition) Blitzer

The solution set for the provided system of equations is $\left\{ \left( -3,\ 4,\ -2 \right) \right\}$.
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 3 & 4 & 2 \\ 4 & -2 & -8 \\ 1 & 1 & -1 \\ \end{matrix} \right|\begin{matrix} 3 \\ -4 \\ 3 \\ \end{matrix} \right]$ Solve this matrix as below to get, $\left[ \left. \begin{matrix} 3 & 4 & 2 \\ 0 & 6 & 4 \\ 1 & 1 & -1 \\ \end{matrix} \right|\begin{matrix} 3 \\ 16 \\ 3 \\ \end{matrix} \right]$ $By,\ 4{{R}_{3}}-{{R}_{2}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 3 & 4 & 2 \\ 0 & 6 & 4 \\ 0 & 1 & 5 \\ \end{matrix} \right|\begin{matrix} 3 \\ 16 \\ -6 \\ \end{matrix} \right]$ $By,\ {{R}_{1}}-3{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 3 & 4 & 2 \\ 0 & 6 & 4 \\ 0 & 0 & 26 \\ \end{matrix} \right|\begin{matrix} 3 \\ 16 \\ -52 \\ \end{matrix} \right]$ $By,\ 6{{R}_{3}}-{{R}_{2}}\to {{R}_{3}}$ Convert the matrix into equation form, \begin{align} 3x+4y+2z=3 & \\ 6y+4z=16 & \\ 26z=-52 & \\ \end{align} Solve the last equation to obtain the value of $z$: \begin{align} & 26z=-52 \\ & z=-\frac{52}{26} \\ & z=-2 \end{align} Substitute the value of $z$ in $6y+4z=16$ to obtain the value of $y$: \begin{align} & 6y+4\times \left( -2 \right)=16 \\ & 6y-8=16 \\ & 6y=24 \\ & y=4 \end{align} Substitute the value of $y,z$ in $3x+4y+2z=3$ to obtain the value of $x$: \begin{align} & 3x+4\times 4+2\times -2=3 \\ & 3x+16-4=3 \\ & 3x=-9 \\ & x=-3 \end{align} So, the solution is $\left\{ \left( -3,\ 4,\ -2 \right) \right\}$ .