Determine whether $ (A_1,A_2,A_3,A_4)$ span $M_{22}$ Determine whether $\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right], \left[\begin{array}{cc} 1&1\\1&1\end{array}\right], \left[\begin{array}{cc} 1&0\\0&1\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ spans $M_{22}$. If it does, construct the set of basis from the spanning set.
Pick an arbitrary matrix in $M_{22}=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$.If we find the following system to be consistent, then this set will span $M_{22}$
$a_1\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right]+ a_2\left[\begin{array}{cc} 1&1\\1&1\end{array}\right]+a_3 \left[\begin{array}{cc} 1&0\\0&1\end{array}\right]+a_4\left[\begin{array}{cc}0&1\\1&0\end{array}\right]=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$ which simplifies to
$$-a_1+a_2+a_3=x$$
$$a_1+a_2+a_4=y$$
$$a_1+a_2+a_4=z$$
$$-a_1+a_2+a_3=w$$
By visually inspecting, we can see that eq #1 & 4 are the same and eq#2 & 3 are the same. So, if we perform row operations, we will end up with an inconsistent system. So, the set do no span $M_{22}$ and we cannot construct the basis.
Am  I on the right track? Appreciate your feedback.

Even if the set doesn't span $M_{22}$, it span some subspace $\subseteq M_{22}$; we need to find the basis for such subspace.
To find it you can proceed as for ordinary row/colum vectors by RREF (in this way you could solve also part 1 by this) and selecting matrices corresponding to the pivot rows.
Notably, collecting the matrices by rows with the entries ordered as for $x,y,z,w$, we have to reduce the following
$$\begin{bmatrix}-1&1&1&-1\\1&1&1&1\\1&0&0&1\\0&1&1&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\1&1&1&1\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$$
from here we can conlude that the subspace spanned has dimension 2 and that we can take the first two matrices as a basis for it.
