Question

In Problems 15–22 determine whether the given set of functions is linearly independent on the interval$\left(‐\infty ,\infty \right)$.

${\mathbf{f}}_{\mathbf{1}}\left(x\right)\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{}{\mathbf{f}}_{\mathbf{2}}\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}{\mathbf{f}}_{\mathbf{3}}\left(x\right)\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}$

Short answer

The set of functions is linearly dependent in the interval $\left(-\infty ,\infty \right)$.

Step-by-step solution

Define Wronskian of the function

Consider each of the functions ${\mathrm{f}}_1\left(\mathrm{x}\right),{\mathrm{f}}_2\left(\mathrm{x}\right),...,{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)$possesses at least$\mathrm{n}-1$ derivatives, then the determinant $\mathrm{W}\left({\mathrm{f}}_1,{\mathrm{f}}_{1,...,}{\mathrm{f}}_1\right)=\left|\begin{array}{cccc}{\mathrm{f}}_1& {\mathrm{f}}_2& ...& {\mathrm{f}}_{\mathrm{n}}\\ {\mathrm{f}}_1^{\text{'}}& {\mathrm{f}}_2^{\text{'}}& ...& {\mathrm{f}}_{\mathrm{n}}^{\text{'}}\\ ⋮& ⋮& ⋮& \mathrm{M}\\ {{\mathrm{f}}^{\left(\mathrm{n}-1\right)}}_1& {{\mathrm{f}}^{\left(\mathrm{n}-1\right)}}_1& ...& {{\mathrm{f}}^{\left(\mathrm{n}-1\right)}}_1\end{array}\right|$, where the primes denote derivatives is known as the Wronskian of the functions.

Find the Wronskian of the set of functions

The given set of functions is:

$$\begin{gathered} {\mathrm{f}}_1\left(\mathrm{x}\right)=\mathrm{x} \\ {\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{x}}^2 \\ {\mathrm{f}}_3\left(\mathrm{x}\right)=4\mathrm{x}-3{\mathrm{x}}^2 \end{gathered}$$

Using the definition of the Wronskian of the function,

$$\begin{gathered} \mathrm{W}\left({\mathrm{f}}_1,{\mathrm{f}}_2,{\mathrm{f}}_3\right)=\left|\begin{array}{ccc}{\mathrm{f}}_1& {\mathrm{f}}_2& {\mathrm{f}}_3\\ {\mathrm{f}}_1^{\text{'}}& {\mathrm{f}}_2^{\text{'}}& {\mathrm{f}}_3^{\text{'}}\\ {\mathrm{f}}_1^{\text{'}\text{'}}& {\mathrm{f}}_2^{\text{'}\text{'}}& {\mathrm{f}}_3^{\text{'}\text{'}}\end{array}\right| \\ =\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& 4\mathrm{x}-3{\mathrm{x}}^2\\ 1& 2\mathrm{x}& 4-6\mathrm{x}\\ 0& 2& -6\end{array}\right| \end{gathered}$$

Find the determinant of the matrix

Determining the determinant of the matrix described by the Wronskian,

$$\begin{gathered} W\left(f_1,f_2,f_3\right)=x\times \left|\begin{array}{cc}2x& 4-6x\\ 2& -6\end{array}\right|-x^2\times \left|\begin{array}{cc}1& 4-6x\\ 0& -6\end{array}\right|+\left(4x-3x^2\right)\times \left|\begin{array}{cc}1& 2x\\ 0& 2\end{array}\right| \\ =x\left[\left(2x\right)\left(-6\right)-\left(4-6x\right)\left(2\right)\right]-x^2\left[\left(1\right)\left(-6\right)-\left(4-6x\right)\left(0\right)\right]+\left(4x-3x^2\right)\left[\left(1\right)\left(2\right)-\left(2x\right)\left(0\right)\right] \\ =x\left(-12x-8+12x\right)-x^2\left(-6\right)+\left(4x-3x^2\right)\left(2\right) \\ =-8x+6x^2+8x^2-6x^2 \\ =0 \end{gathered}$$

Since the value of the determinant is zero and the Wronskian of the set of functions is zero,

Therefore, the set of functions is linearly dependent.