Linear algebra problem, can anyone help?

Originally posted by Lord Shark
Given a system of equations [b]Ax = b, with det(A) = 0, show that there are no solutions unless TRyb = 0, where y is any nontrivial solution of the system TRAy = 0

Since superscript is not available I've used TRA to mean the transpose of matrix A, similarly for TRy.[/b]

Ok, just some comments
a) there is nothing in the description, that A is a SQUARE matrix, and for the given problem there is no need, that A is a square matrix, all of it is true for nXm matrices
b) "a system of equations" or "one equation" just depends of your point of view. If you look at it as single equations, its "a system of", if you look at it as a equation in the vector space, its a single equation. In most questions of linear algebra it is easier to look at the vector space
c) There are a lots of "not"s and "unless"es in the question. If you translate it (removing all the nots (knots?): "there are no solution unless any y..." is equivalent to "if for all y... then there is no solution". And that is the question I talked about.
The triple negation is a stylistic device, which non-omission you should avoid (sorry for my english)
d) sorry, but it looked like a homework
e) and my name is "afx" not "aTx", but of course I am a zero in maths ;-)

Originally posted by afx
Ok, just some comments
a) there is nothing in the description, that A is a SQUARE matrix, and for the given problem there is no need, that A is a square matrix, all of it is true for nXm matrices
b) "a system of equations" or "one equation" just depends of your point of view. If you look at it as single equations, its "a system of", if you look at it as a equa ...[text shortened]... ork
e) and my name is "afx" not "aTx", but of course I am a zero in maths ;-)

Hint: You're not impressing anybody and you're making a fool of yourself.