How did $e^x$ get equal to $e^{2x}$?

Question

$$ \begin{vmatrix} e^x \, \cos(2x) & e^x \, \sin(2x) \\ e^x \, \cos(2x)- 2 e^x \, \sin(2x) & e^x \, \sin(2x)+ 2 e^x \, \cos(2x) \\ \end{vmatrix} $$

I am finding Wronskian examples to work with and in this particular. I'm getting a different answer which is $2e^{2x}$. However, the example used $2e^x$. I rechecked it and it still gives me the same answer. I don't know where i went wrong somewhere I know this is basic for some but if $2e^{2x}$ is correct then is there an explanation for this? Can $e^x$ be used in place of $e^{2x}$? Thank you.

Answer 1

The determinant is linear in its rows (or its columns). This means that you can extract a common factor of each single row or each single column. For a scaled matrix this gives one factor for each row, that is $$ \det(cA)=c^n\det(A) $$ when $A$ is a $n\times n$ matrix. Here $n=2$ and $c=e^x$ with no other exponential terms in the remaining matrix, so that the result has a factor $e^{2x}$.

Answer 2

The Wronskian determinant is

$e^x\cos(2x)(e^x\sin(2x)+2e^x\cos(2x))-e^x\sin(2x)(e^x\cos(2x)-2e^x\sin(2x)$

$=2e^{2x}\cos^2(2x)+2e^{2x}\sin^2(2x)=2e^{2x}$.

Note that $e^xe^x=e^{x+x}=e^{2x}$.