Can the given vector be expressed as a linear combination of the other three Four vectors from $\mathbb{R}^4$ are given: 
$$v_1 = (1, -2, 0, 1), v_2 = (0, 1, -1, -3), v_3 = (1, 0, -2, -5), v_4 = (-3, 5, 1, 0)$$
Can the vector $v_4$ be expressed as a linear combination of the other three? The solution given is yes: $$v_4 = av_1 + (2a + 5)v_2 + (-a - 3)v_3 \qquad a ∈ \mathbb{R}$$
I tried to solve the system of linear equations $\alpha v_1 + \beta v_2 + \gamma v_3 = a v_4$, but this seems to get me nowhere. How can I express $v_4$ as a linear combination of the other three?
The general solution is:

$$ av_1+(5+2a)v_2+(-3-a) v_3 = v_4$$

for any $a \in \mathbb{R}$, as stated. Here's why...

Firstly write $\alpha v_1+\beta v_2+\gamma v_3 = v_4$ out as

$$
\left(\begin{matrix}\alpha+\gamma \\-2\alpha+\beta \\-\beta-2\gamma \\ \alpha-3\beta-5\gamma\end{matrix}\right) = 
\left(\begin{matrix}-3 \\5 \\1 \\ 0\end{matrix}\right)
$$

(I've obviously swapped your row vectors for column vectors.)

If a solution exists for the whole vector, then a solution must exist for its first three elements, so we might as well just work with the following to start with:

$$
\left(\begin{matrix}\alpha+\gamma \\-2\alpha+\beta \\-\beta-2\gamma \\\end{matrix}\right) = 
\left(\begin{matrix}-3 \\5 \\1 \end{matrix}\right)
$$

Re-writing this as a matrix equation, we have:

$$
\left(\begin{matrix}1 &amp; 0 &amp; 1 \\-2 &amp; 1 &amp; 0 \\0 &amp; -1 &amp; -2 \\\end{matrix}\right)
\left(\begin{matrix}\alpha \\ \beta \\ \gamma \\\end{matrix}\right) = 
\left(\begin{matrix}-3 \\5 \\1 \end{matrix}\right)
$$

The determinant of the matrix here is $0$, so we may put $\alpha=a$ where $a \in \mathbb{R}$ is an arbitrary constant. Continuing to solve, we find $\beta=5+2a$ and $\gamma=-3-a$. These values also work to make $\alpha-3\beta-5\gamma=0$ (the fourth element of the original vector equation), so we have the solution!
In your set: $\;v_4\;$ is a linear combination of the other three vectors iff when we form the corresponding matrix with $\;v_4\;$ the last row and we reduce it by rows, this last row becomes all zeros (why?! This must be clear in order to be sure what is going on here...):

$$\begin{pmatrix}
1&amp;-2&amp;0&amp;1\\
0&amp;1&amp;-1&amp;-3\\
1&amp;0&amp;-2&amp;-5\\
-3&amp;5&amp;1&amp;0
\end{pmatrix}\longrightarrow\begin{pmatrix}
1&amp;-2&amp;0&amp;1\\
0&amp;1&amp;-1&amp;-3\\
0&amp;2&amp;-2&amp;-6\\
0&amp;-1&amp;1&amp;3
\end{pmatrix}$$

...and we're done as we can see the last row is the third one multiplied by $\;-\frac12\;$ .