linearly independent and determinant

Question

This question says a matrix $\begin{bmatrix}a & b\\c & d\end{bmatrix}$ where $a_{ij}$ are real numbers. I need to prove that $\det|A|=ad-bc\neq0 \iff $the columns are linearly independent.
Suppose we already know that A's columns are linearly independent iff the system of equations $Ax=0$ where $x=(x_1, x_2)'$ only has trivial solution, that is, $(x_1=0, x_2=0)$ is the only solution.
The following is my answer:
$ \begin{equation} \begin{alignedat}{4} ax_1 &+ bx_2 &&= 0 \\ cx_1 &+ dx_2 &&= 0 \end{alignedat} \end{equation} $ implies $ \begin{equation} \begin{alignedat}{4} adx_1 &+ bdx_2 &&= 0 \\ bcx_1 &+ bdx_2 &&= 0 \end{alignedat} \end{equation} $
Then substract the second equation from the first equation, we get $ x_1(ad-bc)=0$
$(\Rightarrow)$ The columns are linearly independent, so $x_1=x_2=0$ is the only solution. Therefore, (ad-bc) cannot be $0$, otherwise $x_1$ can be any number.
$(\Leftarrow)$ As $ad-bc\neq 0$, we know the only solution to the above equation is $x_1=0$, and hence $x_2=0$. Therefore, the columns are linear independent.
My tutor says the argument is not sufficient, but I can't understand. Could you see the problem?

Answer 1

After you see (hopefully) where you went wrong, I propose the following:

$$\binom ac\;,\;\;\binom bd\;\;\text{linearly dependent}\;\iff\;\exists\;k\in\Bbb R\;\;s.t. \binom bd=k\binom ac\iff$$

$$\binom bd=\binom{ak}{ck}\iff \det A=\begin{vmatrix}a&ak\\c&ck\end{vmatrix}=k\begin{vmatrix}a&a\\c&c\end{vmatrix}=k\cdot 0=0$$

and you're done.