Use thedefinitionof "linearly dependent"! Two vectors, x1 and x2, are "linearly independent" if and only if there exist numbers, A and B, at least one non-zero, such that Ax1+ Bx2= 0. Here, that means A(a, b)+ B(c, d)= (Aa+ Bc, Ab+ Bd)= (0, 0) which is equivalent to Aa+ Bc= 0, Ab+ Bd= 0. Obviously, one solution is A= B= 0. What must be true so that is NOT the only solution? One way to answer that is to try to solve the equations! If we multiply the first equation by b, Aab+ Bbc= 0, multiply the second equation by a, Aab+ Bad= 0, and subtract it from the previous equation, we eliminate A: B(ad- bc)= 0.

IF ad- bc is not 0, we can divide both sides by it getting B= 0. Putting B= 0 into Aa+ Bc= 0, we have Aa= 0. If a is not 0, A=0. If a= 0, we can put B= 0 into the other equation Ab+ Bd= 0 to get Ab= 0. If b is not 0, A= 0 again. (If both A and B are 0, ad- bc= 0.) In any case, the vectors are independent.

If ad- bcis0, then B(ad- bc)= 0 for any value of B- including non-zero values so the vectors are dependent.

In the case of just two vectors we can also say that if ad- bc= 0 and a isnot0, then d= bc/a so that <c, d>= <c, bc/a>= (c/a)<a, b> showing that <c, d> is just amultipleof <a, b>. If a= 0, ad- bc= 0 becomes bc= 0 so either b=0 or c= 0. If b= 0, <a, b>= <0, 0> which is linearly dependent with any other vector. If c= 0, then we have <0, b> and <0, d> so that <0, d>= (d/b)<0, b> and, again, one vector is a multiple of the other.