Prove that the inverse of an integer matrix is also an integer matrix

[bigevil]

Homework Statement

A is an invertible integer matrix. Prove that if det A = 1 or det A = -1, then the inverse of A is also an integer matrix.

Also prove the reverse, if A-inverse is an integer matrix then its determinant is 1 or -1.

Homework Equations

I'm not too sure how to start here. My first instinct is to work backwards from some properties of determinants that I know of.

det A-inv = det A
det [A(A-inv)] = det(A)det(A-inv) = det(A) det(A) = det I = 1, so det A = 1 or -1.

Same can be proven for A-inv, det (A-inv) det (A-inv) = det I = 1, so det A-inv = 1 or -1.

The Attempt at a Solution

I'm not so sure how to approach this from here. Sorry if this is a stupid question, I am studying on my own, not in college and not many other resources to turn to.

 

[HallsofIvy]

One method of finding the inverse of a matrix is to set a^-1_{ij[/b]= Aji/d where Aji is the determinant of the "minor" (matrix you get by removing jth row and ith column) and d is the determinant of A. If A is $\pm 1$, what does that tell you?}

 

[bigevil]

Halls, I'm not sure if I'm familiar with that equation. Do you mean the adjoint matrix of A, ie, A^-1 = Adj (A) / det A? Also not very sure about your question, what do you mean by A is +1 or -1?, the determinant?

If det A = 1 or -1, then the adjoint = inverse.

I have thought of something, not sure if this is the track you're pushing me along:

If det A = 1 or -1, then

A det A = I = A A^-1. Given A is an integer matrix, and already given that det A is an integer, then A^-1 is also an integer matrix. Does that fit together right?

 

[Dick]

Yes, Hall's is talking about Adj(A). It's the transposed matrix of cofactors. If A is integer, then Adj(A) is definitely integer. For the second part, det(I)=det(A)*det(A^(-1)). If A and A^(-1) are integer matrices, then what about det(A) and det(A^(-1))? det(I)=1. Hence?

 

[bigevil]

Ok, Dick:

If A and A-inverse are integer matrices then det A and det (A-inverse) are integers because of the Laplace expansion. det (I) = 1, so det (A A-inv) =1

det (A-inv) det (A-inv) = 1, and since both are equal, det (A-inv) must be 1 or -1.

Is this the right reasoning? Thanx Dick.

 

[MadHawk]

Indeed, the question itself provides the answer. When you say, integer, it means if you have f as a function of integers, then it can be easily manipulated into Drichilet functions and/or Dirac-Delta Functions. And, from here, you can use a lot of identities or transforms to achieve what is desired. Another method is to use induction. Another is to make a 2x2 matrix, where a ij element of I. Then find its inverse. And then prove every element of A inverse, is also an integer. Quite tedious, but has some charming results in the end.

 

[Dick]

Ok, Dick:

If A and A-inverse are integer matrices then det A and det (A-inverse) are integers because of the Laplace expansion. det (I) = 1, so det (A A-inv) =1

det (A-inv) det (A-inv) = 1, and since both are equal, det (A-inv) must be 1 or -1.

Is this the right reasoning? Thanx Dick.

No. Something seems to be evading you. There is NO reason why det(A-inv)*det(A-inv)=1. What you have is det(A)*det(A-inv)=1. det(A) and det(A-inv) are both integers. How many ways are there to multiply two integers and get 1?

 

[bigevil]

Oh... sorry Dick I got it mixed up, was doing a question on determinant of transposes before this.

det A = det (A-inv) = 1 or
det A = det A(-inv) = -1, hence. Thanks Dick...

Madhawk, I'm not much of a mathematician...