Suggested languages for you:

Americas

Europe

Q37E

Expert-verified
Found in: Page 191

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# (M) Show that $$\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}$$ is a linearly independent set of functions defined on $$\mathbb{R}$$. Start by assuming that $${c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)$$Equation (5) must hold for all real t, so choose several specific values of t (say, $$t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}$$) until you get a system of enough equations to determine that the $${c_j}$$ must be zero.

By the inverse matrix theorem, this system has only a trivial solution. Hence, $$\left\{ {t,\sin t,\cos 2t,\sin t\cos t} \right\}$$ is a linearly independent set of functions.

See the step by step solution

## Step 1: Write the given statement

Assume $${c_1} \cdot t + {c_2} \cdot \sin t + {c_3} \cdot \cos 2t + {c_4} \cdot \sin t\cos t = 0$$.

## Step 2: Form a system using specific values of t

The above equation gives a system for  as shown below:

$$\left( {\begin{array}{*{20}{c}}0&{\sin 0}&{\cos 0}&{\sin 0\cos 0}\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right)$$

It means $$Ac = 0$$.

Here, $$A = \left( {\begin{array}{*{20}{c}}0&{\sin 0}&{\cos 0}&{\sin 0\cos 0}\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right)$$.

## Step 3: Find the determinant of A

$$\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}0&0&1&0\\{.1}&{\sin .1}&{\cos .2}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\cos .4}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\cos .6}&{\sin .3\cos .3}\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{.1}&{\sin .1}&{\sin .1\cos .1}\\{.2}&{\sin .2}&{\sin .2\cos .2}\\{.3}&{\sin .3}&{\sin .3\cos .3}\end{array}} \right|\\\det A \ne 0\end{array}$$

## Step 4: Conclusion

By the inverse matrix theorem, the equation $$Ac = 0$$ has only a trivial solution.

Hence, $$\left\{ {t,\sin t,\cos 2t,\sin t\cos t} \right\}$$ is a linearly independent set of functions.