Representing a $2 \times 2$ matrix as a $1 \times 4$ vector? It seems to me (acording to assingment solutions), that you can write a $2 \times 2$ matrix as a column vector instead. Why can you do that?
I just saw a solution to an assignment involving 
$$A_1=\begin{bmatrix}
1&1\\-1&2
\end{bmatrix}$$
Where the matrix was rewritten as $$A_1=\begin{bmatrix}1\\1\\-1\\2\end{bmatrix}$$
 A: If you have $A_1 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then it is false that $A_1 =  \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$.  Those are two different matrices, they are not equal, and if we were to extend the meaning of equality to differently sized matrices, that isn't how it would be done.
But your problem is to determine whether 3 matrices, of size 2 by 2, are linearly independent.  So the first question to ask is, "what does it mean for matrices to be linearly independent" ?
It means that if you sum up scalar multiples of the matrices, you won't get zero.  In other words, there is no such $c$ such that $$\sum_{k=1}^3 c_k A_k = 0$$
Observe that if
$$
\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix} a_3 & b_3 \\ c_3 & d_3 \end{bmatrix}
$$
are linearly independent, then so are
$$
\begin{bmatrix} a_1 \\ b_1 \\ c_1 \\ d_1 \end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix} a_2 \\ b_2 \\ c_2 \\ d_2 \end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix} a_3 \\ b_3 \\ c_3 \\ d_3 \end{bmatrix}
$$
This does not mean that they are equal.  It just means that the linear independence of the first 3 is equal to the linear independence of the last 3.
Analogously, if I said to you "The sum of $x+1$, $x+2$, and $x+3$ is equal to the sum of $x$, $x$, and $x+6$", that does not mean that $x+1 = x$, $x+2=x$, and $x+3 = x+6$.  Same principle, but using linear indpendence instead of sum.
So your problem becomes one of finding the linear independence of 3 vectors, which is a more familiar problem.
A: As @gitgud said, any two vector spaces of the same dimension are isomorphic. Since $M_{2\times 2}(\mathbb{R}) \simeq \mathbb{R}^4$, we may write any $2\times 2$ matrix using any basis of $\mathbb{R}^4$ by the change of basis transformation. 
In your example, the bases are simply the standard bases of $M_{2\times 2}(\mathbb{R})$ and $\mathbb{R}^4$, so the change of basis transformation is essentially the identity, but we have to re-write the matrix a little.
