Linear combination (vectors in space) First of all, sorry if my question is too easy for you guys, and sorry for my por English.. I have serious trouble with vectors haha Can someone please help me?
Given the vectors $$\vec{u} = 4\vec{i}+\vec{j}-3\vec{k}$$ $$\vec{v} = 3\vec{j}+\vec{k}$$ $$\vec{w} = 2\vec{j}+3\vec{k}$$
Justify why $\vec{v}$ can't be expressed as a linear combination of $\vec{u}$ and $\vec{w}$
Thanks!
 A: Taking $$\left|\begin{matrix} 4&1&-3\\0&3&1\\0&2&3\end{matrix}\right|=28\neq 0$$ we find that the this matrix has full rank, i.e. dimension 3.  Hence the rows are independent.
A: Suppose $3j+k=v=\alpha u+\beta w=\alpha(4i+j-3k)+\beta(2j+3k)$. By equation the coefficients of $i$, $j$ and $k$ on the left and the right you get the system of linear equations: $$0=4\alpha,\quad 3=\alpha +2\beta,\quad 1=-3\alpha+3\beta.$$ Is the there a solution of this system? What can you conlude?
Of course, this answer makes sense only if the vectors $i,j,k$ are linearly independent.
A: A linear combination of $\vec{u}$ and $\vec{w}$ would be a vector of the form
$$a\vec{u}+b\vec{w}=a(4\vec{i}+\vec{j}-3\vec{k})+b(2\vec{j}+3\vec{k})=(4a)\vec{i}+(a+2b)\vec{j}+(-3a+3b)\vec{k},$$
where $a$ and $b$ are some numbers. Keeping in mind that 
$$c_1\vec{i}+c_2\vec{j}+c_3\vec{k}=d_1\vec{i}+d_2\vec{j}+d_3\vec{k}$$
if and only if $c_1=d_1$, $c_2=d_2$,  $c_3=d_3$, let's say for the sake of contradiction that there were numbers $a$ and $b$ such that $a\vec{u}+b\vec{w}=\vec{v}$, i.e., 
$$(4a)\vec{i}+(a+2b)\vec{j}+(-3a+3b)\vec{k}=3\vec{j}+\vec{k}.$$
What would $a$ have to be? What seems wrong about $b$, then?
A: Hint: If we write $\vec v = a \vec u + b \vec w $  try to show that $a=0$ ...
