Homogeneous System: Why Invertible A Has No Non-Zero Solutions?

[cupu]

Hello,

In a book I'm reading about linear algebra it's mentioned that in order for the homogeneous system
Ax = 0
to have a solution (other than the trivial solution) the coefficient Matrix must be singular.
The thing is, I can't remember (the wikipedia page on homogeneous systems didn't turn up anything) why if A is invertible, then the system does not have non-zero solutions.

Any help on why this is so is appreciated.

Thank you in advance

 

[Mark44]

If A is invertible, then A^-1 exists.
Ax = 0 ==> A^-1Ax = A^-10 = 0 ==> x = 0

Another way to look at it is that, for A^-1 to exist, both it and A must be one-to-one. The equation Ax = 0 obviously has at least one solution, x = 0, but the one-to-oneness prevents it from having any additional solutions.

 

[cupu]

That makes sense, thank you very much for the response Mark44.