The question asks: Find the "coordinates" of
$v=\begin{bmatrix} -2 & -2 \\ -2 & 4 \end{bmatrix}$
relative to the ordered basis, $F=(f_1, f_2, f_3, f_4)$ where
$f_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, f_2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, f_3 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, f_4 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
I am aware I am supposed to solve this problem using elimination, but at the moment, I am stuck.
I tried to transform these vectors into something I am familiar with. So, I transformed each vector to 4x1 making
$v = \begin{bmatrix} -2 \\ -2 \\ -2 \\ 4 \end{bmatrix}$
and I tried to extend it to each of the $f$ vectors, but elimination forces me to use row exchanges with $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \\ 1 & -1 & 0 & 0 \end{bmatrix}$, the columns corresponding to $f_1, f_2, f_3, f_4$ respectively.
At this point, I'm fairly certain that this problem cannot be this difficult to solve so I'm wondering where I did something wrong. I'm assuming the transformation of the 2x2 matrices into 4x1 is the primary culprit, but I cannot think of an alternative method to solving this problem. Is there a simpler way?
Notice that your ordered basis is in fact an orthogonal basis under the standard inner product. In particular, that means we have $$v = \frac{\langle v,\ f_1\rangle}{2} f_1 + \frac{\langle v,\ f_2\rangle}{2} f_2 + \frac{\langle v,\ f_3\rangle}{2} f_3 + \frac{\langle v,\ f_4\rangle}{2} f_4$$ This gives you $$v = f_1 - 3f_2 -2f_3$$ so your coordinates are $$[v]_F = \begin{pmatrix}1\\-3\\-2\\0\end{pmatrix}$$