Consistency of $A x = b$ and the rank of the augmented matrix

Question

Suppose $A \in \mathbb R^{m \times n}$ and $b \in \mathbb R^m$. What has to be true about the two numbers, $\mbox{rank} \left( [A \, b]\right)$ and $\mbox{rank} \left(A\right)$ in order for the equation $A x = b$ to be consistent?

Here is my attempt.

I know that in order for every $b$ in $R^m$ to be consistent with the equation $Ax=b$, the rank of A must be equal to $n$, in other words, there must be NO zero rows in the row reduced echelon form of $A$. I believe the same logic applies with [$A$ b]

any help will be appreciated.

Answer 1

It is alright for RREF of $A$ to have zero rows, for example $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $b= \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is consistent.

Guide:

To be consistent, $b$ must be in the column space of $A$, that is $b \in \operatorname{span}(A_1, \ldots, A_n\}$.

Can you answer the question about the rank of the two matrix with the information above?

Answer 2

Yes you are right for consistent solution the determinant should not be zero according to rank nullity theorem as nullity is equal to zero and rank is equal to dim of given matrix.So rank is n and given matrix is consistent. Ax=b then x= b$A^{-1}$ this condition satisfy when detA is not equal to zero. Then inverse exist. Overally for consistent rank=n

Answer 3

They must be equal. If, after row reduction, there are any more nonzero entries in the augmented part than the rank of $[A]$, it's inconsistent. Otherwise consistent...