Linear Algebra matrix $Ax=b$ true or false nullspace $Ax=b$
$m$ number of Rows
$n$ number of columns
true or false
A) If $n > m$, given any $b$ you can always solve $Ax=b$.
The answer: False. Counterexample: A is the zero matrix.

We have unknowns more than equations, so we can always solve $Ax=b$. Why the answer is false?
  And even if $A$ is the zero matrix we have the matrices x = zeros then $b$ zeros so there is always answer!!
  Can anyone explain why the answer is false? How can he proved false by $A$ is the zero matrix?

B) If $n < m$, the only solution of $Ax=0$ is $x=0.$
The answer: False. Counterexample: let $A$be the zero matrix.

I understand why it is false but I'm wondering, is there any special case makes this statement true?

Edit:
I made a matrix $A= 2$ rows $\times 1$ column contains $(1,0)$ and assume Ax=0 as the statement, then the only solution is when x=0 because we can't use row echelon form any more. is what I did correct to make the statement true?
 A: For A), the statement is that given any $b \in \mathbb{R}^n$, there is a solution to the matrix equation $Ax = b$.  But if $A$ is the zero matrix, then there is only one $b$ for which there is a solution, namely $b=0$.  Since there are some $b$ for which $Ax = b$ cannot be solved, the original statement is false.
For B), the condition that the only solution to $Ax=0$ is $x=0$ means that there are no free variables in the reduced row echelon form of $A$.  Or said differently, when you apply Gauss-Jordan elimination, you must get $n$ pivots (or equivalently, $n$ non-zero rows).  Said still another way, the only solution to $Ax = 0$ is $x=0$ exactly when the rank of $A$ is equal to $n$.
A: Let $f$ the canonical linear application associated to the matrix $A$ hence
$$f\colon \mathbb{R}^n\to \mathbb{R}^m$$
A) given any $b$ we can solve $f(x)=b$ if $f$ is surjective i.e. $\mathrm{rank}(f)=\mathrm{rank}(A)=m$
B) the only solution to $f(x)=0$ is $x=0$ if $f$ is injective i.e. $\ker(f)=\ker(A)=\{0\}$ 
so by the Rank-nullity theorem $\mathrm{rank}(f)=\mathrm{rank}(A)=n$
