Answer
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 $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.
Work Step by Step
(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 $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.
