Quote:
Originally Posted by Dr Sardonicus
No, sir. If M is a 2x2 matrix, scalar multiplication of M by k produces a matrix with determinant k^2*det(M). Thus, if M is a nonsingular 2x2 matrix, det((1/det(M))M) is 1/det(M).
If M is 2x2 and det(M) is not a square, no scalar multiple of M will have determinant 1.

A peaceful, early morning, especially for Dr Sardonicus,
let: p=31, u1=2, v1=3; so that the norm (u1,v1)=u1²+v1²=13=12⁻¹ mod 31 and 13²=20⁻¹ mod 31
Is it possible from linear algebra to calculate one belonging rotation matrix from this vector ?
The calculated target is:
13*
(27,2)* (2)=(5)
(2,27) (3)=(9)
13*
(14,17)* (2)=(4)
(17,14) (3)=(11)
13*
(24,8)* (2)=(6)
(8,24) (3)=(15)
http://localhost/devalco/unit_circle/system_tangens.php
(the red coloured boxes right, all calculations checked and it seems to be all right.)
My first try:
Let p=31
Let M1=13*M1*=
13*
(a*, b*)
(b*, a*)
1. with det (M1)=1=det (13² * det (M1*)) so that det (M1*)=20 mod 31, therefore a*²+b*²=20
2. with M1*(u1,v1)=(u2,v2) with norm (u1,v1)=(u2,v2)=u1²+v1²=u2²+v2²=13 mod p
so that
13*
(a*, b*) = (u2)
(b*, a*) (v2)
This is more a fragment and should point in one direction, and as it is too late for me,
I hope that some one could finish the calculation.