Answer
The solution set is $\left\{ \left( -2,1,4,3 \right) \right\}$
Work Step by Step
We have the system of equations,
$\begin{align}
& w-x+2y-2z=-1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& x-\frac{1}{3}y+z=\frac{8}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
& y-z=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \\
& z=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{IV} \right)
\end{align}$
Substitute the value of z from the 4th equation to the 3rd equation.
So,
$\begin{align}
& y-z=1 \\
& y-3=1 \\
& y=4
\end{align}$
Now, put the value of $y=4$ and $z=3$ into equation (II) to find the value of x:
$\begin{align}
& x-\frac{1}{3}y+z=\frac{8}{3} \\
& x-\frac{1}{3}\left( 4 \right)+3=\frac{8}{3} \\
& x-\frac{4}{3}+3=\frac{8}{3} \\
& x-\frac{4}{3}+\frac{9}{3}=\frac{8}{3}
\end{align}$
Finally,
$\begin{align}
& x=\frac{8}{3}-\frac{5}{3} \\
& x=1 \\
\end{align}$
Now, put the value of $x=1$ , $y=4$ and $z=3$ into equation (I) to find the value of w:
$\begin{align}
& w-x+2y-2z=-1 \\
& w-1+2\left( 4 \right)-2\left( 3 \right)=-1 \\
& w-1+8-6=-1 \\
& w+1=-1
\end{align}$
Finally,
$\begin{align}
& w=-1-1 \\
& w=-2 \\
\end{align}$
Hence, the solution set is $\left\{ \left( -2,1,4,3 \right) \right\}$.