# Solve a system of linear equations?

GROUPS:
 Hi!I have a pretty easy problem where I have 6 equations and 6 unknown of the form m*x = 0, where m is a 6x6 matrix and x is a vector of length 6 with the 6 unknown variables. I'm looking for a nontrivial solution to this problem so I use the Nullspace command.My code is the following m = {{E^(-I*x*b/2), E^(I*x*b/2), -E^(-I*y*b/2), -E^(I*y*b/2), 0, 0}, {I*x*E^(-I*x*b/2), -I*x*E^(I*x*b/2), -I*y*E^(-I*y*b/2), I*y*E^(I*y*b/2), 0, 0}, {0, 0, -E^(I*y*b/2), -E^(-I*y*b/2), E^(I*x*b/2), E^(-I*x*b/2)}, {0, 0, -I*y*E^(I*y*b/2), I*y*E^(-I*y*b/2), I*x*E^(I*x*b/2), -I*x*E^(-I*x*b/2)}, {E^(-I*x*(a + b/2)), E^(I*x*(a + b/2)), 0, 0, 0, 0}, {0, 0, 0, 0, E^(I*x*(a + b/2)), E^(-I*x*(a + b/2))}} ns = NullSpace[m] This gives me the output { }. What does this result mean? All the coefficients x,y,a,b are real and greater than zero. Is that something that I should specify in the code? I wonder what is wrong with my code? I've attached my code to this post.Thanks in advance! Attachments:
Answer
10 months ago
3 Replies
 Perhaps MatrixForm[RowReduce[m]] and NullSpace[IdentityMatrix[6]] might give a hint
Answer
10 months ago
 For m *x == 0 you must have Det[ m ] == 0, but In[17]:= Det[m] // FullSimplify Out[17]= E^(-I (2 a x + b y)) (-(x + E^(2 I a x) (x - y) + y)^2 + E^(2 I b y) (x - y + E^(2 I a x) (x + y))^2) 
Answer
10 months ago
 No nulls is good nulls?
Answer
10 months ago