Let $(a,b), (c,d)$ be two vectors in $\mathbb R^2$. If $ad-bc=0$ prove that they are linearly dependent Let $(a,b), (c,d)$ be two vectors in $\mathbb R^2$. If $ad-bc=0$ prove that they are linearly dependent
My attempt: I tried to do it by contradiction: Suppose that they are linearly independent that is:
$\alpha_1(a,b)+\alpha_2(c,d)=\vec0$ then $\alpha_1=\alpha_2=0$; solving the system:
$\begin{cases}
\alpha_1a+\alpha_2c=0 \\[2ex]
\alpha_1b+\alpha_2d=0
\end{cases}$
$$\alpha_1ba-\alpha_1ba+\alpha_2bc-\alpha_2da=0 \Rightarrow \alpha_2(bc-ad)=0$$ hence either $bc-ad=0$ or $bc-ad\neq0$ if it is the latter one then we have a contradiction
I would like you to tell me if this is correct :)
 A: If $ad-bc=0$, then $ad=bc$, so $a\cdot(c,d)=(ac,ad)=(ac,bc)=c\cdot(a,b)$.  Therefore $$a(c,d) - c(a,b) = 0.$$
This proves linear dependence unless $a=c=0$.  But if $a=c=0$, then you have 
$$b(0,d) - d(0,b) =0.$$
This proves linear dependence unless also $b=d=0$.  If $a=b=c=d=0$, then $$5(a,b)+72(c,d)=0.$$
A: If you know what are matrices and determinants, you can use them as well.
Say you have two (row) vectors $(a, b)$ and $(c, d)$ and put them in a matrix, one below the other, like this:
$$M := \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$
Then $\det M = ad - bc$, which is zero if and only if $M$ is singular, which is true if and only if its rows (our original vectors) are linearly dependent.
A: Your approach seems correct to me but the writing seems a little ambiguous.
You can improve the writing: Let $\alpha_1(a,b)+\alpha_2(c,d)=0$, without making your assumption about linear independence.
Proceed as you did. In the last step, $\alpha_2(bc-ad)=0$, use your given fact that $bc-ad=0$ to prove that $\alpha_2\neq 0$ which proves linear dependence.
A: We need to prove that $\alpha(a,b)+\beta(c,d)=(0,0)$ have solutions for scalars $\alpha, \beta$ different of zero.
Then, solving for $\alpha, \beta $ you will have $\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\cdot \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Since a homogeneus system of equations have unique or infinite solutions, your unique option is to have infinite number of solutions and this occurs when $\text{rank}\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) < 2$ or equivalent $\det\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc = 0$.
So the solutions for $\alpha, \beta$ are infinite when $ad-bc=0$, that is, is linearly dependent.
You can generalize this for any eucliden vector of $\mathbb{R}^n$.
