## Precalculus (6th Edition) Blitzer

The solution for the provided system of equations is $\left\{ \left( z-2,\ z-2,\ z \right) \right\}$.
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 5 & -11 & 6 \\ -1 & 3 & -2 \\ 3 & -5 & 2 \\ \end{matrix} \right|\begin{matrix} 12 \\ -4 \\ 4 \\ \end{matrix} \right]$ Solve this matrix as below: $\left[ \left. \begin{matrix} 5 & -11 & 6 \\ -1 & 3 & -2 \\ 0 & 4 & -4 \\ \end{matrix} \right|\begin{matrix} 12 \\ -4 \\ -8 \\ \end{matrix} \right]$ $By,\ 3{{R}_{2}}+{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 5 & -11 & 6 \\ 0 & 4 & -4 \\ 0 & 4 & -4 \\ \end{matrix} \right|\begin{matrix} 12 \\ -8 \\ -8 \\ \end{matrix} \right]$ $By,\ 5{{R}_{2}}+{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 5 & -11 & 6 \\ 0 & 4 & -4 \\ 0 & 0 & 0 \\ \end{matrix} \right|\begin{matrix} 12 \\ -8 \\ 0 \\ \end{matrix} \right]$ $By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ 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 drop it. \left[ \left. \begin{align} & \begin{matrix} 5 & -11 & 6 \\ \end{matrix} \\ & \begin{matrix} 0 & 4 & -4 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 14 \\ -8 \\ \end{matrix} \right] Convert this system into linear equations. \begin{align} 5x-11y+6z=12 & \\ 4y-4z=-8 & \\ \end{align} Write the value of x and y in terms of z and solve the equation, \begin{align} & x=z-2 \\ & y=z-2 \end{align} Hence the numbers satisfying the set of solutions $\left\{ \left( z-2,\ z-2,\ z \right) \right\}$