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