If the value of a $2\times2$ determinant is $0$ then one row is multiple of the another

Question

If the value of a determinant is $0$, then a row of the determinant is scalar times of the other row.

I assume that it's true. Let's see a 2x2 determinant $ \begin{vmatrix} a & b \\ c & d \end{vmatrix}=0 $ which is equal to $ ad-bc=0$ and this is $ad=bc$, $a=\frac{bc}{d},d=\frac{bc}{a}$ where $a,d$ are not equal to zero and $c=\frac{ad}{b}$ and $b=\frac{ad}{c}$ so $b,c$ are not equal to zero. How can I finish this proof?

Answer 1

If $a = b = 0$ then clearly $(a,b) = 0(c,d)$ is a scalar multiple of $(c,d)$ so assume that $ab \neq 0$. Then we want to find $t$ such that

$$ (c,d) = t(a,b) \iff c = ta \wedge d = tb. $$

If $a \neq 0$, take $t = \frac{c}{a}$. Then $c = ta$ by definition and $tb = \frac{c}{a} b = \frac{cb}{a} = \frac{ad}{a} = d$ by the fact that the determinant is zero ($ad - bc = 0 \iff ad = bc$).

If $b \neq 0$, take $t = \frac{d}{b}$ and proceed similarly.

Answer 2

Hint: Rewrite $$ad=bc\iff \frac{a}{b}=\frac{c}{d}=:\lambda$$

(Assuming that $b\neq 0 \neq d$. In these cases the left hand side trivializes and you get again get similar results)