Proving type of solutions for matrix
So the question is:
Consider a system of linear equations written in matrix form as Ax =b, where A is a m×n–
matrix, x is an n–dimensional vector–column, and b is an m–dimensional vector–column.
Show that the system has
(a) no solutions if and only if rank A < rank [A|b], where [A|b] is the augmented matrix;
(b) a unique solution if and only if rank A = rank [A|b] = n;
(c) infinitely many solutions if and only if rank A = rank [A|b] < n.
I think I more or less know what Im doing but im confused by one thing.
What I first did was assume that the matrix was in reduced row echelon form and took it from there.
For parts B and C could the x value be zero for a given variable but have the b value be a non-zero number. Wouldnt this still maintain that the rank was equal but there be no solutions. Ill try and explain with an example:
A given row in matrix A is
1 0 0 0 0
X is given by [0,x,x,x,x]
b for this particular row is a non zero value.
Wouldnt this still maintain the rank however make it so there is no solution?
Also is there a way to solve this problem mathematically? My proof has pretty much been all words which I dont find to be a very concrete answer.