## Precalculus (6th Edition) Blitzer

a) The system represented is \begin{align} & w+y+z=0 \\ & w-x+2y+3z=0 \\ & 3w-2x+5y+7z=0 \end{align} b) There are infinitely many solutions existing for this problem.
(a) The system represented by A can be written by multiplying $\left[ A \right]\times \left[ X \right]=\left[ 0 \right]$ Where A is augmented matrix and X is variable matrix. So, \left[ \begin{matrix} 1 & 0 & 1 & 1 & 0 \\ 1 & -1 & 2 & 3 & 0 \\ 3 & -2 & 5 & 7 & 0 \\ \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+y+z=0 \\ & w-x+2y+3z=0 \\ & 3w-2x+5y+7z=0 \end{align} Hence, the system represented is, \begin{align} & w+y+z=0 \\ & w-x+2y+3z=0 \\ & 3w-2x+5y+7z=0 \end{align} (b) Now since $rref\left( A \right)=\left[ \begin{matrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{matrix} \right]$ So, from here we get, \begin{align} & x-y-2z=0 \\ & w+y+z=0 \end{align} We have two equations and three unknowns. So there are infinitely many solutions for this system.