Consistency of the system $AX=b$ In the concept of consistency (compatibility) of a matrix, when $b \in \mathbb R (A)$ we know for sure that the system is consistent or there may exists no solution for it? But when $b \notin \mathbb R (A)$ there is no solution and it is inconsistent. Am I right? 
By $ \mathbb R (A)$, I mean Range of $A$.
 A: I don't know the notation $\mathbb R(A)$, but I think your question is the following:
Let $A \in \mathbb{R}^{m \times n}, b \in \mathbb{R}^m$, we are looking for solutions of the system $$Ax=b$$
Let us denote by $$im(A):=\{y\in \mathbb{R}^m:\exists x \in \mathbb{R}^n \text{ with } Ax = y\}$$ the image of $A$, and $$\ker(A):=\{x \in \mathbb{R}^n: Ax = 0\}$$ its kernel (see). Then:


*

*if $b \notin im(A)$ the system has no solutions which is obvious by definition of $im(A)$.

*if $b \in im(A)$ there exists some $x_0 \in \mathbb{R}^n$ such that $Ax_0 = b$ and the set of all solutions of the system is $\{x_0+v : v \in \ker(A)\}$ (see (1)). It follows that if $\ker(A)=\{0\}$ then the solution $x_0$ is unique (see (2)) and if not then the system has infinitely many solutions (see (3)).


Summary: Every system of linear equations has exactly $0$, $1$ or infinitely many solutions.
Proofs:
(1) Let us show that if $b \in im(A)$ and  $S$ is the set of all solutions of the system $Ax=b$ then for every $x_0 \in S$ holds $S=\{x_0+v : v \in \ker(A)\}$. So let $x_0 \in S$ and $x' \in S$ then $x' = x_0+(x'-x_0)$ and $A(x'-x_0) = b-b = 0$ which shows $S\subset \{x_0+v : v \in \ker(A)\}$. Now for every $v \in \ker(A)$,  we have $A(x_0+v)=Ax_0+Av = b+0 = b$, i.e. $x_0+v \in S$ which implies $S\supset \{x_0+v : v \in \ker(A)\}$.
(2) Suppose $b \in im(A),\ker(A) = \{0\}$ and let $s_1,s_2$ be such that $As_1 = b = As_2$. Then $A(s_1-s_2) = b-b = 0$ and thus $s_1-s_2 \in \ker(A)$. It follows from $\ker(A) = \{0\}$ that $s_1-s_2 = 0$, i.e. $s_1 = s_2$ and thus the solution is unique.
(3) Suppose $b \in im(A),\ker(A) \neq \{0\}$ and let $x_0$ be such that $Ax_0 = b$. Let $v \in \ker(A), v \neq 0$, then we also have $A(\lambda v) = \lambda Av = 0$ for every $\lambda \in \mathbb{R}$. Now, note that $s_\lambda := x_0+ \lambda v$ is a solution of the system $Ax = b$ for every $\lambda \in \mathbb{R}$, because $As_\lambda = A(x_0+\lambda v) = Ax_0 + \lambda Av = b+0=b$. It follows that the system has infinitely many solutions.
