This question was in my exam and I argued with my friends about it.
True or false? Let $A$ be an $m \times n$ matrix of $\operatorname{rank}(A) = m$, the system $Ax=b$ the system has a unique solution for every $b$ that belongs to $\Bbb R^m$.
I think it has a unique solution since the rank of $A$ is $m$ and the $b$ is in $\Bbb R^m$, so there will be $m$ independent columns that can span the $\Bbb R^m$ space. My friends say that they are infinitely many solutions due to the extra $n$. Others say it needs more clarification to the $b$.
The linear system $Ax=b$ has solution if and only if $\text{rank}(A)=\text{rank}(A,b)$.
When $m=n$, $A$ is $n\times n$ square matrix, then $\text{rank}(A)=n$ means it's invertible, multiply $A^{-1}$ on both sides will get $x=A^{-1}b$, it is the unique solution.
When $m>n$, you will find $\text{rank}(A)=m>n$, however the maximum of column rank($=$ row rank) cannot be greater than $n$, so the situation cannot happen.
When $m<n$, the homogeneous equation $Ax=0$ have $n-\text{rank}(A)=n-m$ independent basis to form the solution space of $Ax=0$, the particular solution to $Ax=b$ plus the basis of $Ax=0$ is the whole solution of $Ax=b$, of course there're infinite solutions for every $b$.