I had one doubt in matrix form of linear equations
Say , we have a system of equations $Ax = b$ such that $A$ is an $n\times n$ matrix and $b$ is a $n\times 1$ matrix and so is $x$ then, if we are told that $\operatorname{rank}(A) = n$, then, do we need to check that $\operatorname{rank}(A|b)=n$ or can we say that since $b$ is a linear combination of the component vectors in $A$ then augmenting it in $A$ won't increase the number of linearly independent column vectors and hence the $\operatorname{rank}(A|b)=n$ and can't ever be $n+1$.
I am not sure that I understand you when you write “do we need to check”. Assuming that $\operatorname{rank}(A)=n$, then $\operatorname{rank}(A|b)=n$ if and only if the system $Ax=b$ has a solution. Does this answer your question?