A useful practice: how to determine the number of solutions of a system of linear equations beforehand

We have a system of n linear equations incorporated by Ax = b, with A being a n x p matrix of numeric coefficients, x being a p-dimensional column vector of variables and b being a numeric n-dimensional column vector. A solution of the system is provided by any combination of values in x that makes the system equality true (so it is given by the number of solution vectors x that satisfy the equation above). In order to establish the number of plausible combinations before diving into solving the equations, we can make use of the Rouché-Capelli theorem, which allows us to exploit the notion of rank of a matrix, aka the maximum number of linearly independent row/column vectors of a matrix (what do we mean by that?). We recall that for a generic m x n matrix, its maximum possible rank is min(m,n). Hence, we understand that by augmenting any matrix with one or more columns/rows, we do not lower its rank (why?). This point is crucial since by considering A and A | b (A augmented with b), we state some facts through the theorem.

If rank(A) < rank(A | b), then the system does not admit any solution for x. Indeed, in this case, b would not belong to the column space of A, i.e. we cannot find any combination x that makes Ax = b true. On the other hand, if rank(A) = rank(A | b), then solutions are admitted because b is part of the column space of A. However, we have to make some distinctions here. For instance, restricting ourselves on A being a square matrix (p = n, number of variables equals number of equations), then we would have a unique solution vector if and only if rank(A) = n, in which case the solution is x = A^-1 b (how to find the inverse?). Otherwise, rank(A) < n and we would end up with infinitely many vector solutions, quantified as infinity to the power of n minus rank(A) (this simply means we will have n - rank(A) free variables). In case A is not square, we will definitely have infinitely many solutions when n < p because A will not be full column rank, whereas the reversed scenario n > p should be addressed to check for additional redundant or linearly combined equations. 

Why do we want to do this preliminary check before solving a system using, e.g., the Gauss Jordan elimination method straightforwardly? Essentially to save us time in case we realise, say, that the system is not solvable at all! Whiteboard might be used for illustration and clarification purposes.