# Prove that an nxn matrix is invertible

• Oct 7th 2011, 09:23 AM
GreenyMcDuff
Prove that an nxn matrix is invertible
Afternoon all,

The question:

Let $A=\begin{bmatrix}1&x_1&x_1^2&...&x_1^{n-1}\\1&x_2&x_2^2&...&x_2^{n-1}\\1&.&.&...&.\\1&x_n&x_n^2&...&x_n^{n-1}\\\end{bmatrix}$ where $x_1,x_2,...,x_n$ are distinct real numbers. Show that $A$ 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 $AB=BA=I$ where $B$ is a unique nxn matrix.

So looking at my old uni notes I have:

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

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

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

Thanks

Chris
• Oct 7th 2011, 11:09 AM
Deveno
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.