Originally Posted by

**HallsofIvy** Look at the simple case of $\displaystyle e^{x}$ and $\displaystyle e^{2x}$. If $\displaystyle C_1e^x+ C_2e^{2x}= 0$ then, differentiating, $\displaystyle C_1e^xf+ 2C_2e^{2x}= 0$. Set x= 0 in both equations: $\displaystyle C_1+ C_2= 0$ and $\displaystyle C_1+ 2C_2= 0$. Those are clearly independent equations and have only $\displaystyle C_1= C_2= 0$ as solution.

But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is $\displaystyle \left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1$ not 0.

They are saying that if **any** system of equations has more than one solution, then the determinant of the coefficients must be 0 (if it were not, the matrix of determinants would have an inverse). The point here is to show that the determinant of the coefficients is NOT 0 so there is only the trivial solution.

In general you have

$\displaystyle \left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|$

and you want to show that, as long as all of the numbers $\displaystyle a_i$ are distinct, that is NOT 0.