Linear independence of matrices $I, A, A^2$ I want to prove that $I,A,A^2\:$matrices $\in M_{2\times 2}$ are $\textit {linearly independent}$. 
I consider the following matrices and their "corresponding" vectors:
$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\to (1,0,0,1)$
$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\to (a,b,c,d)$
$A^2=\begin{pmatrix} a^2+ b c & ab+db \\ ac+dc & d^2+bc\end{pmatrix}\to (a^2+bc, ab+db, ac+dc,d^2+bc)$
What I am trying to do, is to prove that these vectors are $\textit{linearly independent}$. 
For that direction, I want to prove that (at least) one $3\times 3$ determinant is $\neq 0$. 
It seems that all $3\times 3$ determinants $\left( e.g. \begin{vmatrix} 1 & 0 & 0\\ a & b & c\\ a^2+bc & ab+db & ac+dc\end{vmatrix} \right)$ are equal to zero! What am I really missing? 

The thing you're trying to prove is not true. Consider the case $A = I$. 
More generally, if $A$ is any $2 \times 2$ matrix, then you can compute its characteristic polynomial, $c(x)$, the determinant of $A - xI$. That will be some quadratic polynomial in $x$, like
$$
3x^2 - 22 x^1 + 8x^0.
$$
It turns out that if you plug in $A$ for $x$ , and treat the $A^0$ as $I$, you'll get a $2 \times 2$ matrix...that's the zero matrix! That's called the Cayley-Hamilton theorem. 
As an example: 
$$
A = \begin{bmatrix}1 & 3 \\ -1 & 1 \end{bmatrix} \\
A- xI = \begin{bmatrix}1-x & 3 \\ -1 & 1-x \end{bmatrix} \\
c(x) = det(A - xI) = (1-x)^2 + 3 = x^2 -2x + 4
$$
Now plug $A$ into $c$ to get
$$
c(A) = A^2 - 2A + 4I \\
 = \begin{bmatrix} -2 & 6 \\ -2 & -2 \end{bmatrix} -  \begin{bmatrix}2 & 6 \\ -2 & 2 \end{bmatrix}  +  \begin{bmatrix}4 & 0 \\ 0 & 4 \end{bmatrix} \\
= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix} 
$$
