Answer
No, by theorem 4.(d and a).
Work Step by Step
Row reducing A,
$\left[\begin{array}{llll}
1 & 3 & 0 & 3\\
-1 & -1 & -1 & 1\\
0 & -4 & 2 & -8\\
2 & 0 & 3 & 1
\end{array}\right]\left(\begin{array}{l}
.\\
+\mathrm{r}_{1}.\\
.\\
-2\mathrm{r}_{1}.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 3 & 0 & 3\\
0 & 2 & -1 & 4\\
0 & -4 & 2 & -8\\
0 & -6 & 3 & -7
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
+2\mathrm{r}_{2}.\\
+3\mathrm{r}_{2}.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 3 & 0 & 3\\
0 & 2 & -1 & 4\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 5
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
\leftrightarrow \mathrm{r}_{4}.\\
.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 3 & 0 & 3\\
0 & 2 & -1 & 4\\
0 & 0 & 0 & 5\\
0 & 0 & 0 & 0
\end{array}\right]\qquad $3 pivot positions...
Row 4 does not have a pivot position.
In Theorem 4,
$\mathrm{d}. A$ has a pivot position in every row,
is false. So then, is
$\mathrm{a}$. For each $\mathrm{b}$ in $\mathbb{R}^{m}$, the equation $A\mathrm{x}=\mathrm{b}$ has a solution
(false)
