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Chapter 4: Problem 92

Show that \(u=(a, b)\) and \(v=(c, d)\) in \(K^{2}\) are linearly dependent if and only if \(a d-b c=0\).

Short Answer

Expert verified
Vectors \(u = (a, b)\) and \(v = (c, d)\) are linearly dependent if and only if their determinant is zero (i.e., \(ad - bc = 0\)). If the determinant is zero, then there exists a scalar k such that \(u = kv\), and vice versa.

Step by step solution

01

Definition of linear dependence

Two vectors are linearly dependent if one can be expressed as a scalar multiple of the other, or in other words, there exists a scalar k such that \(u = kv\). In our case we can write: \[ (a, b) = k(c, d) \]
02

Assume linear dependence and find the determinant condition

Assuming that \(u\) and \(v\) are linearly dependent, let's express their components in terms of the scalar k: \[ \begin{cases} a = kc \\ b = kd \\ \end{cases} \] Divide the two equations: \[ \frac{a}{c} = \frac{b}{d} \] Now we can cross multiply to find the determinant condition: \[ ad = bc \]
03

Assume the determinant condition and show linear dependence

Assuming the determinant condition, i.e., \(ad - bc = 0\), we can rewrite it as \[ ad = bc \] Now, let's find a scalar k such that there exists a scalar k such that \(u = kv\). Divide the components of the vectors: \[ k = \frac{a}{c} = \frac{b}{d} \] Since the scalar k is the same for both components, we have linear dependence between the vectors u and v.
04

Conclusion

We have showed that if the vectors \(u = (a, b)\) and \(v = (c, d)\) are linearly dependent, then their determinant is zero (i.e., \(ad - bc = 0\)). Conversely, we have also shown that if their determinant is zero, then the vectors are linearly dependent. Therefore, \(u\) and \(v\) are linearly dependent if and only if \(ad - bc = 0\).

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Prove that if \(E\) is an elementary matrix, then \(\operatorname{det}\left(E^{t}\right)=\operatorname{det}(E) .\) Visit goo.gl/6ZoU5Z for a solution.

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