Answer
See answer below
Work Step by Step
$Ax=b \\
\rightarrow \begin{bmatrix}
2 & -1 & 1 & 4 \\
2 & 7 & 2 & 3\\
1 & -2 & 5 & 5
\end{bmatrix} \begin{bmatrix}
5 \\
6\\
13
\end{bmatrix} $
We obtain: $M=\begin{bmatrix}
2 & -1 & 1 & 4 | 5\\
1 & -1 & 2 & 3 | 6\\
1 & -2 & 5 & 5 | 13
\end{bmatrix} $
After reducing $M$ we have row-echelor form:
$M=\begin{bmatrix}
1 & -1 & 2 & 3 | 6\\
0 & 1 & -3 & -2 | -7\\
0 & 0 & 0 & 0 | 0
\end{bmatrix} $
Consequently, there are two free variable, $z=t, w=s$. Then $x=-1+t-s, y=-7+3t+2s$
The solution set is: $x=\{(-1+t-s,-7+3t+2s,t,s: t,s \in R\}\\
=\{s(-1,2,0,1)+t(1,3,1,0(-1,-7,0,0): t,s \in R\}$
Hence, $x_p=(-1,-7,0,0)$ is a particular solution of $Ax=b$ and $\{(-1,2,0,1),(1,3,1,0)\}$ forms a basis for nullspace $(A)$.
So all solutions are in the form (4.9.3).
