## Precalculus (6th Edition) Blitzer

Using Gaussian elimination to solve \left\{ \begin{align} 3x+2y+-2z=0 & \\ x-y+z=5 & \\ \end{align} \right. We obtain the matrix \left[ \left. \begin{align} & \begin{matrix} 1 & -1 & 1 \\ \end{matrix} \\ & \begin{matrix} 0 & 1 & -1 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 5 \\ -3 \\ \end{matrix} \right] Translating this matrix back into equation form gives $\left\{ \begin{matrix} \underline{x-y+z=5} & \text{Equation}\ 1 \\ \underline{y-z=-3} & \text{Equation}\ 2 \\ \end{matrix} \right.$ Solving Equation 2 for $y$ in terms of $z$ results in $y=z-3$. Substituting this expression for $y$ in Equation 1 gives $x=2$. The system’s solution set is $\left\{ 2,\ z-3,\ z \right\}$.
Convert the given linear equation into matrix form, \left[ \left. \begin{align} & \begin{matrix} 1 & -1 & 1 \\ \end{matrix} \\ & \begin{matrix} 0 & 1 & -1 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 5 \\ -3 \\ \end{matrix} \right] Write it back in the linear equation, \begin{align} x-y+z=5 & \\ y-z=-3 & \\ \end{align} Write these equations for x and y in terms of z $y=z-3$ and $x=2$ Hence the solution set is $\left\{ \left( 2,\ z-3,\ z \right) \right\}$.