They’ve said
$A*A=I$
$A=\frac{I}{A}$
$A=I*A^{-1}$
$A=A^{-1}$
As inverse of A exists so $|A|\neq 0$
i.e. A is non-singular matrix
if A is non-singular matrix then there is no dependency among rows(like R1=R2+aR3 etc.). And there are no zero rows in the matrix A
also non-singular matrices are full rank matrix(their rank=n)
Now coming to the Augmented matrix (A|B)
as rank of A=n, so rank of the augmented matrix will be n too irrespective of values in B
when rank of A=rank of (A|B)=number of unknown then there exists unique solution