# Proving type of solutions for matrix

• Oct 14th 2009, 02:38 PM
joe909
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.

Thanks
• Oct 15th 2009, 06:40 AM
HallsofIvy
Quote:

Originally Posted by joe909
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]

What is "X"? I see no "X" above.

Quote:

b for this particular row is a non zero value.
What do you mean by "given row"? In row echelon form, only the first row can have the form "1 0 0 0 0" since all columns before the "pivot" (the main diagonal) must be 0.

A matrix has rank less than its size if and only if, in row echelon form, there are one or more rows that consist entirely of 0's. The matrix equation then will have a solution (and, in fact, an infinite number of solutions) if and only if the corresponding rows in the reduce form of b are also 0: That is the case when "rank A = rank [A|b] < n" since "rank A" is the number of non 0 rows in row echelon form and "rank[A|b]" is the number of non 0 rows in the row echelon form of the augmented matrix.

Quote:

Wouldnt this still maintain the rank however make it so there is no solution?
What do you mean by "maintain" the rank"?

Quote:

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.

Thanks
• Oct 15th 2009, 11:36 AM
joe909
Sorry, I realize my original post was not neccessairly clear, I will attempt to clairify.

By "X" i meant "x" as in the n-dimensional matrix mentioned in the problem.

Let me give you an example of my question:

Say the matrix A was

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

Giving it rank = 3

Vector column x = [1,1,0,0]
Vector column b = [1,1,1,0]

That would make the matrix [A|b]

1 0 0 0 | 1
0 1 0 0 | 1
0 0 0 0 | 1
0 0 0 0 | 0

Which also has rank = 3

Therefore it still satisfies Rank A = Rank [A|b] < n
However instead of having unlimited solutions it has none?

This is what is confusing me. Am I doing something wrong?
Your help is very much appreciated, Thanks!