Answer
$\{(1,2,3,-2)\}$.
Work Step by Step
The given system is
$\left\{\begin{matrix}
w & +x & +y &+z &=4 \\
2w& +x & -2y &-z &=0 \\
w& -2x & -y &-2z &=-2 \\
3w &+2x &+y &+3z &=4
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
2 & 1 & -2& -1&0 \\
1&-2&-1&-2&-2\\
3&2&1&3&4
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1,R_3\rightarrow R_3- R_1$ and $R_4\rightarrow R_4-3\times R_1$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
2-2(1) & 1-2(1) & -2-2(1)& -1-2(1)&0 -2(4)\\
1-1&-2-1&-1-1&-2-1&-2-4\\
3-3(1)&2-3(1)&1-3(1)&3-3(1)&4-3(4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & -1 & -4& -3& -8\\
0&-3&-2&-3&-6\\
0&-1&-2&0&-8
\end{array}\right]$
Perform $R_2\rightarrow R_2(-1)$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0(-1) & -1(-1) & -4(-1)& -3(-1)& -8(-1)\\
0&-3&-2&-3&-6\\
0&-1&-2&0&-8
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&-3&-2&-3&-6\\
0&-1&-2&0&-8
\end{array}\right]$
Perform $R_3\rightarrow R_3+3\times R_2$ and $R_4\rightarrow R_4+ R_2$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0+3(0)&-3+3(1)&-2+3(4)&-3+3(3)&-6+3(8)\\
0+0&-1+1&-2+4&0+3&-8+8
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&10&6&18\\
0&0&2&3&0
\end{array}\right]$
Perform $R_3\rightarrow R_3/10$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0/10&0/10&10/10&6/10&18/10\\
0&0&2&3&0
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&1&3/5&9/5\\
0&0&2&3&0
\end{array}\right]$
Perform $R_4\rightarrow R_4-2R_3$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&1&3/5&9/5\\
0-2(0)&0-2(0)&2-2(1)&3-2(3/5)&0-2(9/5)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&1&3/5&9/5\\
0&0&0&9/5&-18/5
\end{array}\right]$
Perform $R_4\rightarrow R_4(5/9)$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&1&3/5&9/5\\
0(5/9)&0(5/9)&0(5/9)&9/5(5/9)&-18/5(5/9)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&4\\
0 & 1 & 4& 3& 8\\
0&0&1&3/5&9/5\\
0&0&0&1&-2
\end{array}\right]$
Convert the matrix into system of equations as shown below.
$\Rightarrow \left\{\begin{matrix}
1w +1x +1y +1z =\;4 \\
0w +1x +4 y +3 z=\;8 \\
0w +0 x +1 y + 3/5 \;z =9/5\\
0 w +0 x +0 y +1 z=\; -2
\end{matrix}\right.$
or we can write.
$\Rightarrow \left\{\begin{matrix}
w +x +y +z =\;4 \\
\; \; \;x +4 y +3 z=\;8 \\
\;\; \;\;\; \;\;\;\; y + 3/5z =9/5\\
\;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; z=\; -2
\end{matrix}\right.$
Substitute back the value of $z$ into third equation.
$\Rightarrow y+(3/5)(-2)=9/5$
Simplify.
$\Rightarrow y-6/5=9/5$
Solve for $y$.
$\Rightarrow y=3$.
Substitute back the value of $z$ and $y$ into the second equation.
$\Rightarrow x +4 (3) +3(-2)=8$
Simplify.
$\Rightarrow x +12 - 6=8$
Solve for $x$.
$\Rightarrow x=2$.
Substitute back the value of $z,y$ and $x$ into the first equation.
$\Rightarrow w +(2) +(3) +(-2) =4$
Simplify.
$\Rightarrow w +2 +3 -2 =4$
Solve for $x$.
$\Rightarrow w=1$.
The solution set is $\{(w,x,y,z)\}=\{(1,2,3,-2)\}$.
