Q:

find equality between linear spans

$$U = Sp\{(2,5,-4,-10), (1,1,1,1),(1,0,3,5), (0,2,-4,-8)\}$$
$$ W = Sp\{(1,-2,7,13), (3,1,7,11), (2,1,4,6) \}$$
two questions:

prove that $U = W$
find the values of the $a \in \mathbb{R}$ where the vector $v=(a,a-6,4a-3,6a-1)$ belongs to $U$

i thought about solving it with a matrix but i wasn't sure how to approach it maybe something like
$$ \begin{pmatrix}
  2 & 1 & 1 & 0 &| &  1 & 3 & 2 \\
  5 & 1 & 0 & 2 &| & -2 &1 &1\\
  -4  & 1  & 3 & -4&|&7& 7& 4  \\
  -10 & 1 & 5 & -8&| & 13 &11 &6
 \end{pmatrix} $$
and to use row reduction some how to reach zero maybe?
also what about the second question was lost there also

A:

Let 
\begin{align*}
A &= 
\begin{bmatrix}
2&1&1&0\\
5&1&0&2\\
-4&1&3&-4\\
-10&1&5&-8
\end{bmatrix}
&
B&=
\begin{bmatrix}
1&3&2\\
-2&1&1\\
7&7&4\\
13&11&6
\end{bmatrix}
\end{align*}
Note the equation $AX=B$ has a (non-unique!) solution
$$
X=
\begin{bmatrix}
-1&-2/3&-1/3\\
3&13/3&8/3\\
0&0&0\\
0&0&0
\end{bmatrix}
$$
Also note the equation $BY=A$ has a (also non-unique!) solution
$$
Y=
\begin{bmatrix}
-13/7&-2/7&1/7&-6/7\\
9/7&3/7&2/7&2/7\\
0&0&0&0
\end{bmatrix}
$$
Do you see how these equations relate to your problem?

 That $AX=B$ has a solution implies that $U\subseteq W$ and that $BY=A$ has a solution implies that $W\subseteq U$. Hence $U=W$.

