Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 4 - Vector Spaces - 4.5 Linear Dependence and Linear Independence - Problems - Page 297: 33

Answer

See answe below

Work Step by Step

We are given: $f_1(x)=\sin (x)$ $f_2(x)=\cos x$ $f_3(x)=\tan x$ and $I=(-\frac{\pi}{2}, +\frac{\pi}{2})$ The augmented matrix of this system is: $W[f_1,f_2,f_3](x)\begin{bmatrix} \sin x & \cos x & \tan x\\ \cos x& -\sin x & \frac{1}{\cos ^2 x} \\ \sin x & -\cos x & \frac{2 \sin x}{\cos ^3x} \end{bmatrix} $ for all $x$ in $(-\frac{\pi}{2}, +\frac{\pi}{2})$ Then $W[f_1,f_2,f_3](\frac{\pi}{4})=\begin{bmatrix} \frac{\sqrt 2}{2}& \frac{\sqrt 2}{2} & 1\\ \frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2}& 2\\ -\frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2} & 4 \end{bmatrix}=\begin{bmatrix} \frac{\sqrt 2}{2}& -\frac{\sqrt 2}{2} \\ -\frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2} \end{bmatrix}-2\begin{bmatrix} \frac{\sqrt 2}{2}& \frac{\sqrt 2}{2} \\ -\frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2} \end{bmatrix}+4\begin{bmatrix} \frac{\sqrt 2}{2}& \frac{\sqrt 2}{2} \\ \frac{\sqrt 2}{2} & -\frac{\sqrt 2}{2} \end{bmatrix}=(-\frac{1}{2}-\frac{1}{2})-2(-\frac{1}{2}+\frac{1}{2})+4(-\frac{1}{2}-\frac{1}{2})=-1-4-0=-5$ Hence, the set of vectors $\{f_1, f_2, f_3 \}$ is linearly independent on $(-\frac{\pi}{2}, +\frac{\pi}{2})$
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