Thread: Prove that an nxn matrix is invertible

Afternoon all,

The question:

Let where are distinct real numbers. Show that 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 where is a unique nxn matrix.

So looking at my old uni notes I have:



The next part is where I get confused because I go on to say that is invertible

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

Thanks

Chris

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.