# Inverse of a matrix

1. Jan 9, 2013

### Artusartos

In this video:

At the time 2:10, I don't understand why we have to note that $$\lambda^3 - 8\lambda$$ and $$\lambda^2 + 1$$ are relatively prime.

Last edited by a moderator: Sep 25, 2014
2. Jan 9, 2013

### rasmhop

The Euclidean algorithm only works for relatively prime elements, so when he says that $r(\lambda), q(\lambda)$ exists such that
$$r(\lambda)(\lambda^3-8\lambda)+q(\lambda)(\lambda^2+1)=1$$
he uses that they are relatively prime.

3. Jan 10, 2013

### Artusartos

Thanks a lot, but...

1) We know that any two elements have a gcd, right? Can't we just use that formula with that gcd?

2) Also, how do we know if they are relatively prime?

4. Jan 14, 2013

### HallsofIvy

Yes, every two elements of an integral domain have a "gcd". They are "relatively prime", by definition, if and only if that gcd is 1. In particular, the "diophantine" equation ax+ by= c has a solution for x and y if and only if any divisor of a and b is also a divisor of c (if n is a divisor of both a and b, a= np, b= nq, then, for any x, y, ax+ by= n(px+ qy) so n divides ax+ by and so must also divide c). In particular, if a and b have a common divisor, so they have a gcd, that also divides c, we can divide the entire equation by it to get a simpler equation in which the coefficients are relatively prime.

We could but, if that gcd is not 1, it is always easier to divide through by gcd to get a simpler equation in which the gcd of the two coefficients is 1- i.e. in which they are relatively prime.

By finding the gcd! Two elements are relatively prime if and only if their gcd is 1.

In your original post one of the elements was $\lambda^3- 8\lambda= \lambda(\lambda^2- 8)$ and it is easy to see that $\lambda^2- 8= 0$ has no integer (or rational) roots so that cannot be factored further. Similarly $\lambda^2+ 1$ cannot be factored with integer (or even real) coefficients. Since they have no factors in common, they are relatively prime.

Last edited by a moderator: Jan 14, 2013
5. Jan 14, 2013

### Artusartos

Thanks a lot.