Answer
See answer below
Work Step by Step
$Ax=b \\
\rightarrow \begin{bmatrix}
1 & 1 & -2 \\
3 & -1 & -7 \\
1 & 1 &1 \\
2 & 2 & -4
\end{bmatrix} \begin{bmatrix}
-3 \\
2 \\
0\\
6
\end{bmatrix} $
We obtain: $M=\begin{bmatrix}
1 & 1 & -2 | -3\\
3 & -1 & -7 | 2\\
1 & 1 &1 | 0\\
2 & 2 & -4 | 6
\end{bmatrix}$
After reducing $M$ we have row-echelor form:
$M=\begin{bmatrix}
1 & 1 & -2 | -3\\
0 & 1 & \frac{1}{4} | -\frac{11}{4}\\
0 & 0 & 1| 1
\end{bmatrix} $
We can notice that there is no free variable.
Hence, $x_p=(2,-3,1)$ is a particular solution of $Ax=b$ and $rank(A)=3$ forms a basis for nullspace $(A)$.
So all solutions are in the form (4.9.3).
