Question

(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.

Short answer

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.

Step-by-step solution

Write the given statement

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

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

Find the determinant of A

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

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 setof functions.