矩阵双模乘法得到标识

Question

我必须乘以矩阵A和B，它可以由数字0,2,3,4,5,6组成，以获得单位矩阵，但是每一步之后的乘积都会与模数发生。例如： -

[A1 A2 A3] and  [B1 B2 B3] 
[A4 A5 A6]    [B4 B5 B6] 
[A7 A8 A9]    [B7 B8 B9] 

((A1*B1)%7+(A2*B4)%7+(A3*B7)%7)%7 = 1 

这将是元素I_11

我怎么能找到两个矩阵A和B？

Answer 1

代码以获得将满足所要求的条件的A和B可能的组合 -

nums = [0,2,3,4,5,6]  

%// allcomb is a MATLAB File-exchange tool available at - 
%// http://www.mathworks.in/matlabcentral/fileexchange/10064-allcomb 
t1 = allcomb(nums,nums) 

t2 = mod(prod(t1,2),7)==1 
out_comb = t1(t2,:) 

输出是 -

out_comb = 

    2  4 
    3  5 
    4  2 
    5  3 
    6  6 

这意味着A和B可能的组合（假设I表示3x3大小的单位矩阵） -

A is 2I, B is 4I and A is 4I, B is 2I %%// 2I would be 2.*I and so on 
A is 3I, B is 5I and A is 5I, B is 3I 
A is 4I, B is 2I and A is 2I, B is 4I 
A is 5I, B is 3I and A is 3I, B is 5I 
A is 6I, B is 6I and A is 6I, B is 6I 

由于指针由@Luis，请注意，你可以混合和匹配这些数字如下拥有的A和B更多的组合选择 -

A as diag([2 4 6]) and B as [4 2 6]) 
A as diag([5 3 2]) and B as [3 5 4]) 

Answer 2

请注意，最后进行模运算就足够了。去除中间模操作不会影响结果。所以条件是mod(A*B,7)应该等于单位矩阵。现在

，因为rem(6^2,7)等于1，一个解决方案是

A = [ 6  0  0 
     0  6  0 
     0  0  6 ]; 

B = [ 6  0  0 
     0  6  0 
     0  0  6 ]; 

检查：

>> rem(A*B,7) 
ans = 
    1  0  0 
    0  1  0 
    0  0  1 

另一种可能性：由于rem(2*4,7)也1：

A = [ 2  0  0 
     0  2  0 
     0  0  2 ]; 

B = [ 4  0  0 
     0  4  0 
     0  0  4 ]; 

，你可以的当然结合：

A = [ 2  0  0 
     0  4  0 
     0  0  6 ]; 

B = [ 4  0  0 
     0  2  0 
     0  0  6 ];