Is it true that in equation $Ax=b$, $A$ is a square matrix of $n\times n$, is having rank $n$, then augmented matrix $[A|b]$ will always have rank $n$?
$b$ is a column vector with non-zero values. $x$ is a column vector of $n$ variables.
If not then please provide an example.
On
Yes. The reason is simple. Let the augmented rows pf $[A|b]$ be $[r_n|b_n]$. A linear dependence in $[A|b]$ means that for some coefficients $c_n$ where all are not zero we have $$c_1[r_1|b_1]+c_2[r_2|b_2]+\cdots+c_n[r_n|b_n]=0$$which means that $$c_1r_1+\cdots+c_nr_n=0\\c_1b_1+\cdots+c_nb_n=0$$and leads to an obvious contradiction so your conclusion is right.
On
Think about the process by which you bring the matrix into row-reduced echelon form. If you’re doing it systematically, you work column by column from left to right. Once you’ve gotten a pivot in a column and cleared the rest of it to zero, the only elementary row operation that you might perform on the matrix after that will have no effect on this column. In other words, nothing to the right of a column has any effect on what that column will look like in the rref. The process terminates when you put a pivot in the last row or all of the remaining rows consist of zeros.
The rref of a full-rank square matrix is the identity. Its last pivot element is in the $n$th row and column, which means that row-reduction stops there. Therefore, you can augment the matrix with as many columns as you like. They don’t affect anything to their left, and the row-reduction process will terminate with an augmented $n\times n$ identity matrix, so that the augmented matrix also has rank $n$.
Yes since we can't have more than $n$ linearly independent vectors the maximum rank for the augmented matrix is $n$.
More in general for a m-by-n matrix we have that rank $\le \max\{m,n\}$.