Prove: The vectors (a, b) and (c, d) in the plane are linearly dependent if and only if ad – bc = 0
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.
To prove the given statement, we need to show that the vectors (a, b) and (c, d) are linearly dependent if and only if ad - bc = 0.
Step 1: Assume that the vectors (a, b) and (c, d) are linearly dependent.
By definition, if two vectors are linearly dependent, then one can be expressed as a scalar multiple of the other.
So, let's assume that the vector (c, d) is a scalar multiple of the vector (a, b). We can represent this as follows:
(c, d) = k(a, b), where k is a scalar.
Step 2: Equate the corresponding components of both vectors:
c = ka
d = kb
Step 3: Solve for k by dividing both equations:
k = c/a
k = d/b
Step 4: Since the two expressions for k are equal, we can equate them:
c/a = d/b
Step 5: Multiply both sides of the equation by ab:
abc = dab
Step 6: Subtract dab from both sides:
abc - dab = 0
Step 7: Factor out a common term:
(a)(bc - ad) = 0
Step 8: Since (a) cannot be zero (otherwise the vectors would be the zero vector and not linearly dependent), we have:
bc - ad = 0
Therefore, we have shown that if (a, b) and (c, d) are linearly dependent, then ad - bc = 0.
Step 9: To prove the other direction (if ad - bc = 0, then (a, b) and (c, d) are linearly dependent):
Assume that ad - bc = 0.
Step 10: Rearrange the equation:
ad = bc
Step 11: Divide both sides by bd (assuming b and d are non-zero):
a/b = c/d
Step 12: Let k = a/b = c/d, where k is a scalar.
Step 13: Rewrite the equation:
a = kb
c = kd
Step 14: The vector (a, b) can then be expressed as a scalar multiple of the vector (c, d) and vice versa, so they are linearly dependent.
Therefore, we have proved that (a, b) and (c, d) are linearly dependent if and only if ad - bc = 0.