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!

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.

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.

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.