To prove the statement, we need to show two things:
1. If the vectors (a, b) and (c, d) are linearly dependent, then ad - bc = 0.
2. If ad - bc = 0, then the vectors (a, b) and (c, d) are linearly dependent.
Proof:
1. If the vectors (a, b) and (c, d) are linearly dependent, it means that there exist scalars k1 and k2, not both zero, such that k1(a, b) + k2(c, d) = (0, 0). Expanding this equation gives:
(k1a + k2c, k1b + k2d) = (0, 0)
This implies that k1a + k2c = 0 and k1b + k2d = 0.
Rearranging the first equation gives k1a = -k2c. Multiplying both sides by d gives k1ad = -k2cd.
Rearranging the second equation gives -k1b = k2d. Multiplying both sides by c gives -k1bc = k2cd.
Adding these two equations together gives: k1ad - k1bc = k2cd - k2cd. Simplifying gives ad - bc = 0. So, if the vectors (a, b) and (c, d) are linearly dependent, then ad - bc = 0.
2. If ad - bc = 0, we need to show that there exist scalars k1 and k2, not both zero, such that k1(a, b) + k2(c, d) = (0, 0).
Since ad - bc = 0, we can rearrange the equation as ad = bc. This implies that a/b = c/d or ad/b = c. Let k1 = d and k2 = -b. Notice that k1 and k2 are not both zero since b is not equal to d, so they satisfy the conditions.
By substituting k1 and k2 into the equation, we get k1(a, b) + k2(c, d) = (da, db) + (-bc, -bd) = (da - bc, db - bd) = (0, 0).
Therefore, if ad - bc = 0, then the vectors (a, b) and (c, d) are linearly dependent.
Since we have proved both directions, we conclude that the vectors (a, b) and (c, d) in the plane are linearly dependent if and only if ad - bc = 0.
Prove: The vectors (a, b) and (c, d) in the plane are linearly dependent if and only if ad – bc = 0
1 answer