Linear system of equations and its solutions I know this is a trivial question, but I still have a little doubt on it... If we consider the system of equations
\begin{cases} 
x + y = 1 \\ 
x + ay = 0 
\end{cases} 
When we consider the RREF of the augmented matrix, we have:
\begin{bmatrix} 
1 & 0 & \frac{a}{a-1} \\
 0 & 1 & \frac{1}{1-a}\\
\end{bmatrix}
From which we know that the system admits unique solution as long as $\det A \neq 0 \iff a \neq 1$ (where $A$ represents the coefficient matrix). However, looking at the RREF of the augmented matrix, this system does not admit a case in which the solutions are infinite. Am I right or not? Thanks for the help :) 
 A: We have $Ax=b$, and for $a\neq 1$ we have $\det(A)\neq 0$, so the unique solution is given in this case by $x=A^{-1}b=\frac{1}{a-1}\begin{pmatrix} a \cr -1 \end{pmatrix}$. So this is exactly what your RREF says. 
The only other case possible is $\det(A)=0$, hence $a=1$. In that case, we have $0=x+y=1$, which is a contradiction. Consequently the case of infinitely many solution can never occur.
A: You are correct that there is no value of $a$ for which the system has infinitely many solutions, but I'm not sure you can see that from your final matrix -- at least, I can't.
When $a = 1$, as you've noted, the determinant of the coefficient matrix is $0$, and so there is not a unique solution. But it's not clear from the final matrix whether there are infinitely many or no solutions. In this case, our augmented matrix is 
$$\left[\begin{array}{cc|c}1 & 1 &1  \\ 1 & 1 &0\end{array}\right].$$
Now when you row reduce, you'll get 
$$\left[\begin{array}{cc|c}1 & 1 &0 \\ 0 & 0 &-1\end{array}\right],$$
with the final row corresponding to the (solutionless!) equation $0 = -1$, and we can see the system is inconsistent.
The point here is that you have to examine values of $a$ (or whatever parameter) that cause the determinant to be $0$ separately; when performing the row reduction on your original matrix, you divide by $0$ several times when $a = 1$, and the resulting matrix is meaningless in these situations.
