How would you show that $Ax=b$ has a solution $\iff \text{rank } A = \text{rank } [A\, b]$? I am wondering what the $[A \,b]$ here stands for, is it matrix and its multiplied result?
So could anybody please give me some hints to start the problem as I only have the intuition that if $Ax=b$ and $x$ is not $0$, then rank $A$ has to be equal to rank $[A\, b]$.
 A: Hint: If $Ax=b$ has a solution, say $x_0$, then $b$ can be written as a linear combination of the columns of $A$ (with coefficients, exactly the coefficients of $x_0$).

Now, by definition a vector (here $b$) is independent of a set of other vectors (here the other columns of $A$) if he cannot be written as a linear combination of vectors in this set. This should help you prove both directions of the if and only if statement.
A: The notation $[A\ b]$ is ad hoc. The author means: the $m\times n$ matrix $A$ followed by the $m\times 1$ matrix $b,$ considered as a single $m\times (n+1)$ matrix.
The rank of this matrix can never be less than the rank of $A$ because it actually contains $A.$ It is equal to the rank of $A$ if and only if the newly added column $b$ is a linear combination of the columns of $A,$ which is just a fancy way of saying that the system has a solution.
A: Perform the Gaussian elimination on $A|b$ obtaining $X|y$. $rank(A) \neq rank(A|b)$ $\iff$ $X$ has fewer non-null rows than $X|y$ $\iff$ $X|y$ has an equation of the form $0=c, c\neq 0$ $\iff$ $Ax=b$ has no solution.
