Answer
$\{(2,1,-1,3)\}$.
Work Step by Step
The given system is
$\left\{\begin{matrix}
w & +x & +y &+z &=5 \\
w& +2x & -y &-2z &=-1 \\
w& -3x & -3y &-z &=-1 \\
2w &-x &+2y &-z &=-2
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
1 & 2 & -1& -2&-1 \\
1&-3&-3&-1&-1\\
2&-1&2&-1&-2
\end{array}\right]$
Perform $R_2\rightarrow R_2- R_1,R_3\rightarrow R_3- R_1$ and $R_4\rightarrow R_4-2 R_1$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
1-1 & 2-1 & -1-1& -2-1&-1-5 \\
1-1&-3-1&-3-1&-1-1&-1-5\\
2-2(1)&-1-2(1)&2-2(1)&-1-2(1)&-2-2(5)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&-4&-4&-2&-6\\
0&-3&0&-3&-12
\end{array}\right]$
Perform $R_3\rightarrow R_3+4\times R_2$ and $R_4\rightarrow R_4+ 3R_2$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0+4(0)&-4+4(1)&-4+4(-2)&-2+4(-3)&-6+4(-6)\\
0+3(0)&-3+3(1)&0+3(-2)&-3+3(-3)&-12+3(-6)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&-12&-14&-30\\
0&0&-6&-12&-30
\end{array}\right]$
Perform $R_3\rightarrow R_3/(-12)$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0/(-12)&0/(-12)&-12/(-12)&-14/(-12)&-30/(-12)\\
0&0&-6&-12&-30
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&1&7/6&5/2\\
0&0&-6&-12&-30
\end{array}\right]$
Perform $R_4\rightarrow R_4+6R_3$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&1&7/6&5/2\\
0+6(0)&0+6(0)&-6+6(1)&-12+6(7/6)&-30+6(5/2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&1&7/6&5/2\\
0&0&0&-5&-15
\end{array}\right]$
Perform $R_4\rightarrow R_4/(-5)$.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&1&7/6&5/2\\
0/(-5)&0/(-5)&0/(-5)&-5/(-5)&-15/(-5)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 1 & 1& 1&5\\
0 & 1 & -2& -3&-6 \\
0&0&1&7/6&5/2\\
0&0&0&1&3
\end{array}\right]$
Convert the matrix into system of equations as shown below.
$\Rightarrow \left\{\begin{matrix}
1w +1x +1y +1z =\;5 \\
0w +1x -2 y -3 z=\;-6 \\
0w +0 x +1 y + 7/6 \;z =5/2\\
0 w +0 x +0 y +1 z=\; 3
\end{matrix}\right.$
or we can write.
$\Rightarrow \left\{\begin{matrix}
w +x +y +z =\;5 \\
\; \; \;\;\;\;x -2 y -3 z=\;-6 \\
\;\; \;\;\; \;\;\;\; y + 7/6z =5/2\\
\;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\; z=\; 3
\end{matrix}\right.$
Substitute back the value of $z$ into third equation.
$\Rightarrow y+(7/6)(3)=5/2$
Simplify.
$\Rightarrow y+7/2=5/2$
Solve for $y$.
$\Rightarrow y=-1$.
Substitute back the value of $z$ and $y$ into the second equation.
$\Rightarrow x -2 (-1) -3(3)=-6$
Simplify.
$\Rightarrow x +2 - 9=-6$
Solve for $x$.
$\Rightarrow x=1$.
Substitute back the value of $z,y$ and $x$ into the first equation.
$\Rightarrow w +(1) +(-1) +(3) =5$
Simplify.
$\Rightarrow w +1 -1 +3 =5$
Solve for $x$.
$\Rightarrow w=2$.
The solution set is $\{(w,x,y,z)\}=\{(2,1,-1,3)\}$.
