 Solution

Write $u=(a,b{)}^T$, $v=(c,d{)}^T$ and $\Delta =ad-bc$. If $xu+yv=0$, thenEliminating $y$ gives $\Delta x=0$, and eliminating $x$ gives $\Delta y=0$. Hence $\Delta \ne 0$ implies linear independence.

For the converse, suppose $\Delta =0$. If $u=0$, the family already contains the zero vector and is linearly dependent. If $a\ne 0$, the equality $ad=bc$ gives $v=(c/a)u$. If $a=0$ but $u\ne 0$, then $b\ne 0$ and $bc=0$ forces $c=0$, so $v=(d/b)u$. Both cases give a nontrivial linear relation. ThereforeThis is the two-dimensional determinant criterion, proved directly without assuming that criterion.