Please help solving Linearly Dependent question for 4 variation

cosx, sinx, e^{x}, and e^{-x} are LI or LD ( 0 =or< x =or< 1)

I can see without calculation that it is LI but I do not know how to prove it.

I tried to use Wronskian Determinant but it gets really messy

W(x) =

cosx sinx e^{x} e^{-x }-sinx cosx e^{x }-e^{x }-cosx -sinx e^{x }e^{-}^{x}

sinx -cosx e^{x }e^{-}^{x }Someone please suggest a better and quicker way...(Headbang)

Thank you :(

Re: Please help solving Linearly Dependent question for 4 variation

Hi angelme,

I will call the Wronskian matrix $\displaystyle W.$ To start, we should have $\displaystyle W_{4,4}=-e^{-x};$ in other words the entry in the fourth row, fourth column should be $\displaystyle -e^{-x},$ not $\displaystyle e^{-x}.$ I'm guessing you actually have this and missed a minus sign in your post.

To make life simpler, using the fact that the determinant is not affected when you add a row to another will make things easier. For example, adding row 1 to row 3 makes row 3 become $\displaystyle 0, 0, 2 e^{x}, 2e^{-x}.$ If you use this fact a couple of times you should be able to get a row that has 3 zeros and one nonzero entry. You then expand your determinant about this nonzero entry and go from there.

Give it another shot with this in mind. I ended up getting $\displaystyle det(W)=-6$ when all was said and done. Since $\displaystyle -6\neq 0$ the vectors are linearly independent. Let me know if you still get stuck. Good luck!

Re: Please help solving Linearly Dependent question for 4 variation

sorry I'm still not getting it :( Could you please explain me step by step?