Find invertible matrix pairs where both matrices are integers

Hello!

I think the summary says it all! Looking for 3x3, but assume that's not critical.

Have found solution for matrix pair where one is integer, but other is not.

Thanks!

Doug Anderson

[James Tursa]

Do you mean matrices where both the matrix and its inverse contain only integer elements? What do you mean by "matrix pair"? Can you give an example? Do you mean something like the case where the determinant is 1 so that the cofactors will give you integer elements for the inverse if the original matrix has only integer elements?

[James Tursa]

The cofactors comment was just by way of understanding the problem ... not a suggested method to use. If you have a matrix M with all integer elements and you know the determinant is 1, then you could probably just use round(inv(M)) as a first cut as long as the sizes of the integers are not too big. Do you already have a method for generating the original matrix of integer elements with determinant 1, or are you asking for that as well?

What is this for? I.e., what is the actual problem you are working on?

[Douglas Anderson]

I have a method to generate a matrix of integer elements, i.e., doubles that look like integers! I know you can't invert a true integer matrix, AND "isinteger" only identifies if it's a true integer class. Is there a method to say that this is an integer "for all practical purposes"?

This is for security software. Have the matrix I posted, just looking for others. It's been working well.

Again, thanks.

Doug

Answer 1

[John D'Errico]

It seems trivial.

A = eye(3)

A = 3×3

1 0 0 0 1 0 0 0 1

inv(A)

ans = 3×3

1 0 0 0 1 0 0 0 1

Both are entirely integer. As it turns out, A is idempotent, so it is its own inverse, but that was not expressed as an issue.

But surely you are asking about other matrices, that are also fully integer, yet have an all-integer inverse? This too should be entirely doable. The trick is to consider how to create them.

N = 3;
M = 5;
L = tril(randi(2*M,[N,N])-M,-1) + eye(N)

L = 3×3

1 0 0 1 1 0 1 -4 1

A = L*L'

A = 3×3

1 1 1 1 2 -3 1 -3 18

format long g
inv(A)

ans = 3×3

27 -21 -5 -21 17 4 -5 4 1

So A is a fully integer randomly created matrix. It has the property of a fully integer inverse. (if we ignore the crap that sometimes arise in the least significant bits of the inverse. They are just due to floating point trash.) If you wish to know why this works, look carefully at how I created the matrix. Look carefully at the matrix L.

A nice property of this creation scheme, is it will work for any size of matrix A. Just change the value of N. As well, it will be blazingly fast, requiring no solve.

[John D'Errico]

An interesting question about an integer valued matrix where the inverse is also integer valued, is if the eigenvalues must all be 1.

Answer 2

[Walter Roberson]

syms v
M = [1 2 3; 4 5 7; 8 9 v]

M = 

sol = solve(det(M) == 1)

sol = 

12

newM = subs(M, v, sol(1))

newM = 

inv(newM)

ans = 

[Douglas Anderson]

Thank you, Walter, but I don't have the symbolic math toolbox!