## Precalculus (6th Edition) Blitzer

a) The system represented is \begin{align} & W+2X+5Y+5Z=-3 \\ & W+X+3Y+4Z=-1 \\ & W-X-Y+2Z=3 \end{align} b) The complete solution of the system is $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.
(a) The system represented by A can be written by multiplication of $\left[ A \right]\times \left[ X \right]=\left[ 0 \right]$ Where A is augmented matrix and X is variable matrix. So, \left( \begin{matrix} 1 & 2 & 5 & 5 & -3 \\ 1 & 1 & 3 & 4 & -1 \\ 1 & -1 & -1 & 2 & 3 \\ \end{matrix} \right)\times \left( \begin{matrix} W \\ X \\ Y \\ Z \\ 1 \\ \end{matrix} \right)=\left( \begin{align} & 0 \\ & 0 \\ & 0 \\ & 0 \\ & 0 \\ \end{align} \right) So, \begin{align} & W+2X+5Y+5Z=-3 \\ & W+X+3Y+4Z=-1 \\ & W-X-Y+2Z=3 \end{align} Hence, the represented system is thus found. (b) Now, $\text{rref}\left( A \right)=\left[ \begin{matrix} 1 & 0 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{matrix} \right]$ So, \begin{align} & X+2Y+Z=-2 \\ & W+Y+3Z=1 \end{align} Now find $W$ and $X$ in terms of $Y$ and $Z$. So, \begin{align} & W+Y+3Z=1 \\ & W=1-Y-3Z \end{align} And, \begin{align} & X+2Y+Z=-2 \\ & X=-2-2Y-Z \end{align} Hence, the complete solution of the system is $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.