Linear equations given by $Ax = b$ have a solution if and only if $rank(A|b) = rank(A)$

Question

Theorem:

Given a system of linear equations $Ax = b$ where $A \in M_{m \times n}\left(\mathbb{R}\right)$, $ x \in \mathbb{R}^{n}_{col} $, $ b \in \mathbb{R}^{m}_{col}$

Deduce that a solution $x$ exists if and only if $rank\left(A|b\right) = rank\left(A\right)$ where $A|b$ is the augmented coefficient matrix of this system

I am having trouble proving the above theorem from my Linear Algebra course, I understand that A|b must reduce under elementary row operations to a form which is consistent but I don't understand exactly why the matrix A|b need have the same rank as A for this to happen.

Please correct me if I am mistaken

Answer 1

To see this note that rank of the matrix is dimension of the span of columns of the matrix. Now if $Ax=b$ has solution, then it means that some linear combination of columns of A gives us b, which implies that $b$ lies in $span(A)$ and so $rank(A|b)=rank(A)$. You can argue similarly in the reverse direction.

Answer 2

To get started,

If the system contains a row such that [ 0 0 0 0 | b ] with b=/=0 then the system is inconsistent and has no solution. The rank is the number of pivots matrix X has in echelon form, whereby b is the pivot in this row.

Let M=[A ,B], the augmented matrix, where A is the original matrix.

The system has a solution if and only if rank(A)=rank(M).

If b is a pivot of M then the solution does not exist. Hence rank(m)>rank(a).