Answer
No, $\mathrm{b}$ is not a linear combination of the columns in B.
No, the columns in B are vectors of $\mathbb{R}^{4}$, not $\mathbb{R}^{3}$.
Work Step by Step
Row reducing B,
$\left[\begin{array}{llll}
1 & 4 & 1 & 2\\
0 & 1 & 3 & 4\\
0 & 2 & 6 & 7\\
2 & 9 & 5 & -7
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
.\\
-2\mathrm{r}_{1}.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 4 & 1 & 2\\
0 & 1 & 3 & 4\\
0 & 2 & 6 & 7\\
0 & 1 & 3 & -11
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
-2r_{2}.\\
-\mathrm{r}_{2}.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 4 & 1 & 2\\
0 & 1 & 3 & 4\\
0 & 0 & 0 & 15\\
0 & 0 & 0 & -7
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
.\\
+\frac{7}{15}\mathrm{r}_{3}.
\end{array}\right)$
$\sim\left[\begin{array}{llll}
1 & 4 & 1 & 2\\
0 & 1 & 3 & 4\\
0 & 0 & 0 & 15\\
0 & 0 & 0 & 0
\end{array}\right]\qquad...$ 3 pivot positions.
Row 4 does not have a pivot position.
In Theorem 4,
$\mathrm{d}. B$ has a pivot position in every row,
is false. So then, is
$\mathrm{a}$. Each $\mathrm{b}$ in $\mathbb{R}^{m}$ is a linear combination of the columns of $\mathrm{B}$.
(false)
