Answer
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 $\left[ \begin{matrix}
-5.5 \\
2 \\
1 \\
0 \\
\end{matrix} \right]$
Work Step by Step
(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 $\left[ \begin{matrix}
-5.5 \\
2 \\
1 \\
0 \\
\end{matrix} \right]$.
