Answer
See answer below
Work Step by Step
$Ax=b \\
\rightarrow \begin{bmatrix}
1 & 3 & -1 \\
2 & 7 & 9\\
1 & 5 & 21
\end{bmatrix} \begin{bmatrix}
4 \\
11\\
10
\end{bmatrix} $
We obtain: $M=\begin{bmatrix}
1 & 3 & -1 | 4\\
2 & 7 & 9 | 11 \\
1 & 5 & 21 |10
\end{bmatrix}$
After reducing $M$ we have row-echelor form:
$\begin{bmatrix}
1 & 3 & -1 | 4 \\
0 & 1 & 11 | 3 \\
0 & 0 & 0 | 0
\end{bmatrix}$
Consequently, there are one free variable, $z=t$. Then $x=-5+34t, y=3-11t$
The solution set is: $x=\{(-5+34t,3-11t,t : t \in R\}\\
=\{t(34,-11,1)+(-5,3,0): t \in R\}$
Hence, $x_p=(-5,3,0)$ is a particular solution of $Ax=b$ and $(34,-11,1)$ forms a basis for nullspace $(A)$.
So all solutions are in the form (4.9.3).
