*Originally posted by afx*

**First of all, Ax = b is not a system of equations, but just one equation.
**

I assume, you are talking about matrices and vectors over a field and not about modules over a ring ( "det A != 0" should be "det A invertible" ).

Ok. Let us have the equation A x = b (det A 0 or not), and lets have an y as solution of ATy = 0 and with yTb != 0 . ( I prefer a postfix " ...[text shortened]... his homework.

By the way, you shoud do your homework at home and not just pasting it here

*Originally posted by afx*
Thanks for your solution. There are some errors that you have made, here are a few:

**First of all, Ax = b is not a system of equations, but just one equation.**
This is false.

**A, b** and

**x** represent a square matrix, a column vector and a column vector respectively. In the context of the question this represents a system of equations in x1, x2, ...xn.

**By the way, you shoud do your homework at home and not just pasting it here**
It isn't my homework and I am not a student, so this rather sanctimonious statement is based on a false premise.

**But if there is a nontrivial solution of ATy = 0, then det A must be zero.
**

So det A = 0 is no precondition, but a concolusion.
The question postulates the existence of such a solution, so I think saying the determinant is zero is just like saying x != 0 when considering y = 1/x. It is just tidy.