Take the 2-minute tour ×  Let $u$ be the $3\times 1$ matrix
$$\begin{bmatrix} 0\\ 4\\ 4 \end{bmatrix}$$  and $A$ the
$3\times2$ matrix: $$\begin{bmatrix} 3 & -5\\ -2 & 6\\ 1 & 1
\end{bmatrix}.$$  Here's where the title comes in: is $u$ in the
plane $\mathbb R^3$ spanned by the columns of $A$? Why or why
not? I know that it is, I just don't know why. This is the
chapter before we learn about linear dependence and
independence, so I doubt it has to do with either of these,
beyond that I haven't a clue.  share|improve this question I
tried $5/3a_1 + a_2$. Then I tried $5a_1 + 3a_2$. It became
obvious. Alternatively, see the answer below where you actually
solve a system to find the linear combination. A nice
alternative is to find the linear combination of $a_1, a_2$ that
gives you $(0, 1, 0)$ and the linear combination that gives you
$(0, 0, 1)$. –  user2468 Sep 4 '12 at 21:53 add comment  3
Answers  Assuming that the column vectors of your matrix $A$ can
be written with $v=\begin{bmatrix}3&-2&1\end{bmatrix}^T$