$x$ and $x^2$ linearly dependent? Consider the functions $x$ and $x^2$ on $\mathbb{R}$. Clearly, they are linearly independent.
But consider the following argument.
Consider the matrix 
$$A = \begin{bmatrix}
x & x^2\\
0 & 0\\
\end{bmatrix}$$
Clearly, the determinant is zero. This implies the existence of a nonzero matrix, $$B =\begin{bmatrix} a\\ b\\ \end{bmatrix}$$ such that $$AB=0$$.
This implies that $ax+bx^2=0$ for some nonzero $a$ or some nonzero $b$. But this implies that $x$ and $x^2$ are linearly dependent.
Clearly, false.

Where’s the flaw?
 A: They are linearly independent as functions of variable $x$. Yes, you are right that given a specific value of $x$ you can find non-zero $a,b\in\mathbb{R}$ such that $ax+bx^2=0$. But you will never find non-zero $a,b$ which will work for all $x\in\mathbb{R}$. 
A: When discussing linear independence of the columns of a matrix, you must allow the coefficients in the linear combinations to come from the same space as the entries in the matrix (otherwise linear algebra doesn't in any way work the way you're used to). And clearly, if we allow $a$ and $b$ to be polynomials, then $ax+bx^2=0$ has nontrivial solutions.
A: $x$ and $x^2$ are considered to be elements of the vectorial space $\mathbb{R}[X]$ or  $\mathbb{R}[X]_2 $ endowed with the basis $1,x,x^2$. Then 
\begin{align*}
x=&0\times 1 + 1\times x+ 0\times x^2\\
x=&0\times 1 + 0\times x+ 1\times x^2
\end{align*}
,hence $\begin{pmatrix} 0 \\ 1\\0\end{pmatrix}$,  $\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}$ are coordinates of $x$ and $x^2$ respectively. Clearly, this two vectors are linearly independents.
