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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 :)

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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.

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    Yes, but I want to know if there are real values of $a$ for which the system can admit infinite solutions... just for curiosity! :) – 2017-01-01
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    Oh, you mean infinitely *many* solutions. That is impossible, of course. – 2017-01-01
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    Exactly, thank you, sorry for my poor english :) – 2017-01-01
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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.