Answer
$\mathrm{u}$ belongs to the plane spanned by the columns of A.
Work Step by Step
We check whether $\mathrm{A}\mathrm{x}=\mathrm{u}$ has a solution.
If it does, then $\mathrm{u}$ is a linear combination of the columns of A.
If so, then u belongs to the plane spanned by the columns of A.
Reduce the augmented matrix $[\mathrm{A}\ \mathrm{x}]$
$\left[\begin{array}{lll}
3 & -5 & 0\\
-2 & 6 & 4\\
1 & 1 & 4
\end{array}\right]\left(\begin{array}{l}
+\mathrm{r}_{2}.\\
.\\
.
\end{array}\right)\sim\left[\begin{array}{lll}
1 & 1 & 4\\
-2 & 6 & 4\\
1 & 1 & 4
\end{array}\right]\left(\begin{array}{l}
..\\
+2\mathrm{r}_{1}.\\
-\mathrm{r}_{1}.
\end{array}\right)\sim$
$\sim\left[\begin{array}{lll}
1 & 1 & 4\\
0 & 8 & 12\\
0 & -8 & -12
\end{array}\right]\left(\begin{array}{l}
..\\
.\\
+\mathrm{r}_{2}.
\end{array}\right)\sim\left[\begin{array}{lll}
1 & 1 & 4\\
0 & 8 & 12\\
0 & 0 & 0
\end{array}\right]\left(\begin{array}{l}
..\\
\div 8.\\
.
\end{array}\right)\sim$
$\sim\left[\begin{array}{lll}
1 & 1 & 4\\
0 & 1 & 3/2\\
0 & 0 & 0
\end{array}\right]\left(\begin{array}{l}
-r_{2}.\\
.\\
.
\end{array}\right)\sim\left[\begin{array}{lll}
1 & 0 & 5/2\\
0 & 1 & 3/2\\
0 & 0 & 0
\end{array}\right]$
The equation has a solution, so
$\mathrm{u}$ belongs to the plane spanned by the columns of A.
