It is about the matrix equation AX=B, where a, b c in A and B are independent elements, X should be made of constants.
Case 1
A=[a b 0;0 0 c];
B=[a b 0 0;0 0 c c];
The solution is X=[1 0 0 0;0 1 0 0;0 0 1 1];
It seems to be unique. But how to prove it?
Case 2
A=[a b 0 0;0 0 c c];
B=[a b;c c];
The solution is X=[1 0;0 1;0 0;1 1], or [1 0;0 1;1 1;0 0], or [1 0;0 1;1 0;0 1], or [1 0;0 1;0 1;1 0] ;
Apparently, in this case, the solution is not unique. But how to prove it?
This seems to be simple but really confuses me.
The solution in the case1 can not be unique, since there are 8 linear equations for 12 unknowns. It must depend on 4 arbitrary parameters. The same is to the case2.
Victor J. Аdlucky Dear sir, the solution X should be invariant with a, b, c. With such constraints, X is unique in case 1. But I don't know how to prove it.
Dear Liming Yuan,
Observing common 4-parametric solution in the case1 X={{u,v,t,s}, {a*(1-u)/b, 1-a*v/b, -a*t/b, -a*s/b}, {0,0,1,1}} we conclude that solution, invariant with a, b, c, exists if and only if u=1, v=s=t=0, i.e. X={{1,0,0,0}, {0,1,0,0}, {0,0,1,1}}.
Similarly, observing common 4-parametric solution in the case2 X={{t,s},{a*(1-t)/b, 1-a*s/b},{u,v},{1-u,1-v}} we conclude that there exists 2-parametric invariant solution if and only if t=1, s=0, i.e. X= {{1,0}, {0,1}, {u,v}, {1-u,1-v}}.
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