## Precalculus (6th Edition) Blitzer

The solution for the given system of equations is $\left\{ \left( -2z+5,\ -2+z,\ z \right) \right\}$.
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 8 & 5 & 11 \\ -1 & 4 & 2 \\ 2 & -1 & 5 \\ \end{matrix} \right|\begin{matrix} 30 \\ 3 \\ 12 \\ \end{matrix} \right]$ Solve this matrix as below:. $\left[ \left. \begin{matrix} 8 & 5 & 11 \\ -1 & 4 & 2 \\ 0 & -9 & 9 \\ \end{matrix} \right|\begin{matrix} 30 \\ 3 \\ 18 \\ \end{matrix} \right]$ $By,\ 4{{R}_{3}}-{{R}_{1}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 8 & 5 & 11 \\ 0 & -27 & 27 \\ 0 & -9 & 9 \\ \end{matrix} \right|\begin{matrix} 30 \\ 54 \\ 18 \\ \end{matrix} \right]$ $By,\ 8{{R}_{2}}+{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 8 & 5 & 11 \\ 0 & -27 & 27 \\ 0 & 0 & 0 \\ \end{matrix} \right|\begin{matrix} 30 \\ 54 \\ 0 \\ \end{matrix} \right]$ $By,\ {{R}_{2}}-3{{R}_{3}}\to {{R}_{3}}$ Convert the last row into equation form, \begin{align} 0x+0y+0z=0 & \\ 0=0 & \\ \end{align} The last row does not add any information about the variables; therefore we will drop it. \left[ \left. \begin{align} & \begin{matrix} 8 & 5 & 11 \\ \end{matrix} \\ & \begin{matrix} 0 & -27 & 27 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 30 \\ 54 \\ \end{matrix} \right] Convert this system into linear equations. \begin{align} & 8x+5y+11z=30 \\ & -27y+27z=54 \\ \end{align} Now write the value of x and y in terms of z and solving the equation we get, \begin{align} & x=5-2z \\ & y=z-2 \\ \end{align} Hence, the solution for the given system of equations is $\left\{ \left( -2z+5,\ -2+z,\ z \right) \right\}$.