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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.

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    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 $(0, 0, 1)$.2012-09-04

3 Answers 3