Answer
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.
Work Step by Step
(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.
