By calculating the Wronskian, determine whether the set of exponential functions
\[\{e^x, e^{2x}, e^{3x}\}\] is linearly independent on the interval $[-1, 1]$.

Solution.

The Wronskian for the set $\{e^x, e^{2x}, e^{3x}\}$ is given by
\[W(x):=\begin{vmatrix}
e^x & e^{2x} & e^{3x} \\
e^x &2e^{2x} &3e^{3x} \\
e^x & 4e^{2x} & 9e^{3x}
\end{vmatrix}.\]

We compute
\begin{align*}
W(x)&=e^xe^{2x} e^{3x}\begin{vmatrix}
1 & 1 & 1 \\
1 &2 &3 \\
1 & 4 & 9
\end{vmatrix}\\[6pt] &=2e^{6x}.
\end{align*}

Since the Wronskian $W(x)=2e^{6x}$ is never zero, we conclude that the set $\{e^x, e^{2x}, e^{3x}\}$ is linearly independent.

(Note that in general, we just need to show that $W(x_0)\neq 0$ for some point $x_0 \in [0, 1]$.

For example, since we have $W(0)=2\neq 0$, we can conclude that $\{e^x, e^{2x}, e^{3x}\}$ is linearly independent.)

Related Question.

Try the following generalized version.

The solution is given by Exponential Functions are Linearly Independent.

The post Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent first appeared on Problems in Mathematics.