Is $[0,4,4]^T$ in the plane in $\mathbb{R}^3$ spanned by the columns of $A$? 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.
 A: Assuming that the column vectors of your matrix $A$ can be written with $v=\begin{bmatrix}3&-2&1\end{bmatrix}^T$ and $w=\begin{bmatrix}-5&6&1\end{bmatrix}^T$ where $A=[v\quad w]$ then your (linear independant) vectors $v$ and $w$ span obviously one plane $P=\operatorname{span}\{v,w\}\subset\mathbb{R}^3$ with $\dim(P) = 2$.
Assume that there is one vector $x\in P$. $x$ has then the following form 
$$x=\lambda v+\mu w\in P$$
and therefore can be written as a linear combination of the two vectors $v,w$ that span $P$. If you want to check whether $u=\begin{bmatrix}0&4&4\end{bmatrix}^T$ is in $P$ you have to find $\lambda,\mu$ such that the combination of the vectors $v,w$ equals $u$. If there is no solution, then the vector is not part of the plane.
Basically you solve the following system of equations:
$$\begin{bmatrix}3&-5\\-2&6\\1&1\end{bmatrix}\cdot\begin{bmatrix}\lambda\\\mu\end{bmatrix}=\begin{bmatrix}0\\4\\4\end{bmatrix}$$
Hint: This system has a unique solution.
A: You should try to write $u$ as $u = Ax$ when $x$ is a 2d vector. If you can find $x$, then the answer is yes. (What is more difficult is to show that it's impossible when it is actually impossible.)
A: simply compute the rank of the the augmented matrix [A u]. It tells is the number of linearly independent columns. If rank is 2, then u is in the plane spanned by columns of A.
