Prove that an nxn matrix is invertible

Afternoon all,

The question:

Let $\displaystyle A=\begin{bmatrix}1&x_1&x_1^2&...&x_1^{n-1}\\1&x_2&x_2^2&...&x_2^{n-1}\\1&.&.&...&.\\1&x_n&x_n^2&...&x_n^{n-1}\\\end{bmatrix}$ where $\displaystyle x_1,x_2,...,x_n$ are distinct real numbers. Show that $\displaystyle A$ is invertible.

How far I have got:

So there are several ways to do this from what I gather. The way I have tried is by trying to prove that $\displaystyle AB=BA=I$ where $\displaystyle B$ is a unique nxn matrix.

So looking at my old uni notes I have:

$\displaystyle AB=I\implies C(AB)=CI\implies (CA)B=C$

The next part is where I get confused because I go on to say that $\displaystyle CA=I \implies B=C \therefore A$ is invertible

Can anyone explain to me if this is correct and why it is so?

Thanks

Chris

Re: Prove that an nxn matrix is invertible

if CA= I, and AB = I, then B = C:

C =C I = C(AB) = (CA)B = IB = B.

thus B ( = C) is "one inverse" for A. suppose B' is another, so that B'A = AB' = I:

B' = B'I = B'(AB) = (B'A)B = IB = B.

so a matrix that is invertible can only have one inverse, and if a matrix has a left-inverse and a right-inverse, both the left-inverse and the right-inverse

are the same matrix, A^-1.

so to find an inverse for A, you can just find a right-inverse or a left-inverse, which saves time checking by matrix multiplication.

personally, i would check det(A) for this particular problem, because A^-1 exists iff det(A) ≠ 0.