## Precalculus (6th Edition) Blitzer

a) The system represented is; \begin{align} & 2w+17x-23y+40z=0 \\ & 2w+5x+y+3z=0 \\ & x-2y+3z=0 \end{align} b) The solution for this system is $\left[ \begin{matrix} -5.5 \\ 2 \\ 1 \\ 0 \\ \end{matrix} \right]$
(a) 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} 2 & 17 & -23 & 40 & 0 \\ 2 & 5 & 1 & 3 & 0 \\ 0 & 1 & -2 & 3 & 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} & 2w+17x-23y+40z=0 \\ & 2w+5x+y+3z=0 \\ & x-2y+3z=0 \end{align} Hence, the system represented is; \begin{align} & 2w+17x-23y+40z=0 \\ & 2w+5x+y+3z=0 \\ & x-2y+3z=0 \end{align} (b) Now since $rref\left( A \right)=\left[ \begin{matrix} 1 & 0 & 5.5 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{matrix} \right]$ So, from here we get, \begin{align} & w+5.5y=0 \\ & x-2y=0 \\ & z=0 \end{align} So, \begin{align} & w=-5.5y \\ & x=2y \\ & z=0 \end{align} So, the solution will be $\left[ \begin{matrix} -5.5 \\ 2 \\ 1 \\ 0 \\ \end{matrix} \right]$ Hence, the solution to given problem will be $\left[ \begin{matrix} -5.5 \\ 2 \\ 1 \\ 0 \\ \end{matrix} \right]$.