Finding a linear transformation $L$ such that $ L(\vec{u}_{i}) = {v}_{i}$ for $ 1 \leq i \leq 3$. I've been tearing my hair out for hours over this seemingly straightforward linear question. I know it's bad form to post homework without some partial amount of work done, but my work is garbage. I think I'm misunderstanding this question very badly and would appreciate a sanity check.
Define these six vectors: 
$$ \vec{u}_{1} = (1,4,7,2) \qquad \vec{u}_{2} = (3,1,1,5) \qquad \vec{u}_{3} = (-3,10,19,-4) $$ 
$$ \vec{v}_{1} = (7,1,6,5) \qquad \vec{v}_{2} = (-2,1,3,1) \qquad \vec{v}_{3} = (11,3,2,9) $$
Find a linear transformation $L$ such that $ L(\vec{u}_{i}) = {v}_{i}$  for $ 1 \leq i \leq 3$.
(Hint: write down the three equations $L(\vec{u}_{i}) = {v}_{i}$ in terms of the columns of the coefficient matrix of $L$. This will be a system of three equations with four $vectors.$ Find four vectors that satisfy this system.)
 A: It is impossible.
Note that $u_3 = 3 u_1-2 u_2$. Hence we should have $v_3=L(u_3) = 3 L(u_1)-2 L(u_2) = 3 v_1-2 v_2$.
However, it is easy to check that this is not the case, hence no such linear transformation exists.
(Possibly there is a transcription error?)
A: For convenience, let me take the elements of $\mathbb{R}^4$ to be column vectors, and in particular replace the $u_k$'s and $v_k$'s by their transposes, the corresponding column vectors.
Well, let $L : \mathbb{R}^4 \to \mathbb{R}^4$ be your linear transformation, and let $A$ be the matrix of $L$ with respect to the standard ordered basis of $\mathbb{R}^4$; let $A_k \in \mathbb{R}^4$ be the $k$-th column of $A$, $k=1,\dotsc,4$. Then for each $k$, by unpacking the matrix multiplication, one can check that
$$
 v_k = L(u_k) = A u_k = (u_k)_1 A_1 + (u_k)_2 A_2 + (u_k)_3 A_3 + (u_k)_4 A_4,
$$
where $(u_k)_j$ denotes the $j$th entry of $u_k$. Plugging your given $u_k$ and $v_k$ gives the hinted-at system of three linear equations in four unknown vectors.
