# Find the determinant of matrix $n \times n$.

Given the matrix: $$A_{n}=\begin{bmatrix} e^x&e^{2x}& \cdots &e^{nx}\\ e^x&2e^{2x}& \cdots &ne^{nx}\\ \vdots & \vdots & \ddots & \vdots \\ e^x&2^{n-1}e^{2x}& \cdots &n^{n-1}e^{nx}\end{bmatrix}$$ I need to prove that $\det(A_n) \neq 0$.

I've found det for $A_3$ and $A_4$. But how to find det for common case $A_n$? (or at least show that it is not zero)

• maybe with induction and use of Laplace – user409387 Oct 30 '17 at 18:45

The determinant is $$e^{x+2x+\cdots+nx}\left|\matrix{1&1&\cdots&1\\1&2&\cdots&n\\ \vdots&\vdots&\ddots&\vdots\\1&2^{n-1}&\cdots&n^{n-1}}\right|.$$ This is a Vandermonde determinant and is nonzero.
• @НиколайЖурба Write $D$ for $d/dx$. Then $f(D)y=0$ has solutions $e^{a_1x},\ldots,e^{a_nx}$ where $f(t)=(t-a_1)\cdots(t-a_n)$. – Lord Shark the Unknown Oct 30 '17 at 19:20
Hint: Your determinant is the Wronskian (see https://en.wikipedia.org/wiki/Wronskian) of the following set $$\{e^{x},e^{2x},\ldots,e^{nx}\}.$$