Linear Algebra - Determine if the set of matrices in $M_{22}$ is linearly independent or linearly dependent Is the set of matrices $\{\begin{bmatrix}0&1\\0&1\end{bmatrix},\begin{bmatrix}1&1\\2&2\end{bmatrix},\begin{bmatrix}1&4\\2&2\end{bmatrix},\begin{bmatrix}5&4\\2&6\end{bmatrix}\}$ in $M_{22}$ linearly independent or linearly dependent?
For some reason I'm stuck.
What I have done: 
$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = 0$ 
What I was thinking of doing is: 
$A_1(det(A_1)) +A_2(det(A_2) + A_3(det(A_3)) + A_4(det(A_4)) = 0$
I don't know if that's right but I do know I'm on the right track! Any help would be much appreciated!
 A: Looking into the determinant in this way is heading in the wrong direction.  The determinant is not preserved under addition.  To illustrate:
$$
\overbrace{\begin{bmatrix} 2 & 2 \\ -1 & 2 \\ \end{bmatrix}}^{\text{det}=6}
+
\overbrace{\begin{bmatrix} 0 & -1 \\ 1 & -2 \\ \end{bmatrix}}^{\text{det}=1}
=
\overbrace{\begin{bmatrix} 2 & 1 \\ 0 & 0 \\ \end{bmatrix}}^{\text{det}=0}.$$
To continue, notice that the system of equations $$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$$
is equivalent to
\begin{align*}
A_2+A_3+5A_4 &= 0 \\
A_1+A_2+4A_3+4A_4 &= 0 \\
2A_2+2A_3+2A_4 &= 0 \\
A_1+2A_2+2A_3+6A_4 &= 0.
\end{align*}
We can solve this in the usual way (i.e., Gaussian elimination).
Alternatively, we write the matrices as the rows of a $4 \times 4$ matrix
$$
\begin{bmatrix}
0 & 1 & 0 & 1 \\
1 & 1 & 2 & 2 \\
1 & 4 & 2 & 2 \\
5 & 4 & 2 & 6 \\
\end{bmatrix}.
$$
If the rows of this matrix are linearly independent (which occurs if and only if the matrix has non-zero determinant), then the original set of matrices is linearly independent.
A: Hint : You get 4 equations with 4 unknowns if you solve the matrix equation
       elementwise. Note, that the sum is the null-matrix, not the number 0.
You do not need determinants.
A: Your first equation is correct, where $A_1,..A_4$ are scalars.
Note, that equation gives you:
$A_2+A_3+5A_4=0$
$A_1+A_2+4A_3+4A_4=0$
$2A_2+2A_3+3A_4=0$
and $A_1+2A_2+2A_3+6A_4=0$
Equations $1$ and $3$ then imply that $A_4=0$.
Equations $3$ and $4$ imply that $A_1=0$.
Ans then equations $1$ and $2$ imply that $A_2=A_3=0$.
So these are indeed linearly ind.
