I have following:
Proposition Leave $ bf A $ be a $ m $ by $ n $ matrix and $ bf b = O $ then the linear
system $ bf Ax = b $ has a solution $ implies $ $ rank ( mathbf {A mid b}) =
range ( mathbf A) $.
My attempt:
I'll use next $ bf R $ to denote $ rref (A) $.
$ ( rightarrow) $
We test by contradiction.
Suppose $ bf Ax = b $ has a solution and $ range ( mathbf {A mid b}) ≠
range ( mathbf A) $
We are required to consider two cases:

$ rank ( mathbf {A mid b})>
range ( mathbf A) $

$ range ( mathbf {A mid b}) <
range ( mathbf A) $
one)
Suppose $ rank ( mathbf {A mid b})> rank ( mathbf A) $.
$ mathbf {A mid b} $ is row equivalent to $ bf R mid b & # 39; $, Thus $ rank ( mathbf {A mid b}) = rank ( bf R mid b & # 39;) $
$ mathbf {A} $ is row equivalent to $ bf R $, Thus $ rank ( mathbf {A}) = rank ( bf R) $
Therefore we have $ rank ( mathbf {R mid b & # 39;})> rank ( mathbf R) $
It follows that there must be at least one row so that
$$ mathbf {R mid b & # 39;} =
left ( begin {array} {cccc  c}
cdots & cdots & cdots & cdots & cdots \
cdots & cdots & cdots & cdots & cdots \
cdots & cdots & cdots & cdots & cdots \
0 & 0 & cdots & 0 & a \
cdots & cdots & cdots & cdots & cdots \
end {array} right) $$
Where $ to ≠ 0 $.
But that would mean that the linear system is inconsistent. As $ bf R mid b & # 39; $ It is equivalent to $ mathbf {A mid b} $, follow that linear system $ bf Ax = b $ it is inconsistent too, which contradicts the fact that $ bf Ax = b $ it has a solution. Thus $ rank ( mathbf {A mid b}) not>
range ( mathbf A) $
2)
Suppose $ rank ( mathbf {A mid b}) <rank ( mathbf A) $. Leave $ range ( mathbf A) = m $.
$ mathbf {A mid b} $ is row equivalent to $ bf R mid b & # 39; $, Thus $ rank ( mathbf {A mid b}) = rank ( bf R mid b & # 39;) $
$ mathbf {A} $ is row equivalent to $ bf R $, Thus $ rank ( mathbf {A}) = rank ( bf R) $
So, we have $ range ( mathbf {R}) = m $Y $ range ( mathbf {R mid b & # 39;}) <m $
Consider matrix $ ( mathbf {R mid b & # 39;}) $
$$ mathbf {R mid b & # 39;} =
left ( begin {array} {c  c}
mathbf {r} _1 & b_ {1} \
mathbf {r} _2 & b_ {2} \
vdots & vdots \
mathbf {r} _m & b_ {m} \
vdots & vdots \
end {array} right) $$
As $ range ( bf R mid b & # 39;) $ $ <m $, it follows that the row vectors $ bf r $ are linearly dependent and vector $ mathbf {r} _m $ It can be written as the linear combination of the previous vectors. This is a contradiction, since we have said that $ rank ( mathbf {A}) = rank ( mathbf {R}) = m $, which implies that the vectors $ bf r $ It must be linearly independent. Thus, $ range ( bf R mid b & # 39;) $ $ not <m $, and from $ rank ( mathbf {A mid b}) = rank ( bf R mid b & # 39;) $ implies that
$ range ( bf A mid b) $ $ not <m $.
We have shown that both cases are impossible and, therefore, we can conclude that $ rank ( mathbf {A mid b}) =
range ( mathbf A) $. $ Box $
It is right?
Actually, the initial proposition required to show $ iff $, but as you can see, the post would be too long, therefore, I tried to demonstrate $ implies $ First.