/sci/ - Science & Math

Integral 2x2 matrices have integral inverses if and only if they have determinant of 1 or -1. The formula of the determinant of (A + nB) is quadratic in n with the same coefficients for all n (det B n^2 + cross term + det A). You use the pigeonhole principal to give you the existence of 3 quadratic equations with the same right hand sign (+1 or -1 ). From there you take linear combinations to give you 2 quadratics in n with different coefficients and constant terms equal to zero. You conclude that the coefficients, det B and the cross term, must be zero. Plug back in to the equation with n=5 and you see that the determinant is the same as that of A.

>>4357933

This seems right, but I didn't completely follow the second half of your proof so let me explain it in a different way.

Like you said, integral 2x2 matrices have integral inverses if and only if they have determinant of 1 or -1. Therefore, det(A), det(A+B), det(A+2B), det(A+3B), det(A+4B) are all +1 or -1. Furthermore, we know that det(A+nB) is a quadratic polynomial in n.

Now, we know that any nonconstant polynomial of degree at most two takes any real value at most twice, by the fundamental theorem of algebra. But det(A+nB) takes the value +1 or the value -1 at least three times by the pigeonhole principle.

We conclude that det(A+nB) is a constant polynomial. It must be either +1 for all n, or -1
for all n. Either way, det(A+5B) is +1 or -1, so A+5B has an inverse with integer entries.

>>4358815
Thanks for clarifying that, makes a lot more sense now.

>>4357933
>You use the pigeonhole principal

reported for being a retard

>>4359371
You are given 51 problems, only 50 of which can be solved without the pidgeonhole principle
WHAT NOW BITCH

What's the matter with the pigeonhole principle?

>>4359474
Try to stuff pigeons in holes and see what happens.

(A + 5B)^-1 = (A+4B)^-1 + (A+B)^-1 - A^-1 and since you have only added/subtracted integers the result must be an integer.

>>4361143

Nope. Try A = [ 1 0 / 0 1 ], B = [ 0 1 / 0 0 ] and you'll see that your formula doesn't hold.

>>4361143
Why must this equality hold in this case? It doesn't hold for <span class="math">2\times2[/spoiler] matrices in general.

>>4361429
Actually, the formula holds for those particular A and B.

>>4357933
>>4358815
Well done. I didn't see the shortcut with the pigeonhole principle, though it's obvious in retrospect. I used the fact that <span class="math">\det(A+nB)=\pm1[/spoiler] for <span class="math">n=0,\ldots,4[/spoiler] to directly (and inelegantly) show that <span class="math">\det(A+5B)=\pm1[/spoiler].

>>4361768

It's ok; I have Aspergers.