System of equations has no or infinite solutions? For which $ \lambda \in \mathbb{R} $ has the following system of equations
$ x- \lambda y = 1 $
$ (\lambda - 1)x - 2y = 1 $
a unique solution and no solutions in $ \mathbb{R} $?
I solved for $x$ in the first equation and substituted for it in the second, and ended up with this: $ y = - \frac{\lambda+2}{(\lambda+1)(\lambda-2)} $.
Now, I know the interval within which there is a unique solution and that there are no solutions when $\lambda = -1$. However, it is the case when $\lambda = 2$ that confuses me. Would the system have no solutions because of the division by zero or would it have an infinite number of solutions because both equations will be identical?
 A: Infinite number of solutions when $\lambda = 2$. Since both equations are $x-2y = 1$, any $(x,y)$ that satisfies that relation will satisfy both equations, and there are an infinite number of such pairs.
A: It would have an infinite number of solutions. Thats why you should actually instead, solve the system of equations:
$$\begin{bmatrix}1 & -\lambda \\ \lambda - 1 & -2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 \\ 1 \end{bmatrix} $$
Once you perform row reduction, you should get to:
$$\begin{bmatrix}1 & -\lambda \\ 0 & 2 - \lambda(\lambda-1)\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 \\ \lambda - 2 \end{bmatrix} $$
So then we only care about the second row, which now says that:
$$ (2 - \lambda^2 +\lambda)y = \lambda -2  $$ 
Now there are two cases as you correctly point out. If $\lambda = -1$, you get that $0 = -3$ which is a contradiction, so no solutions there.
On the other hand, if $\lambda = 2$, then you get $0y = 0$, this is not a contradiction; but there is also no solution to this problem, which means that there are infinite solutions.
A: 
The system has solutions if and only if the matrix of the homogeneous
  linear system and the augmented matrix have the same rank. When there
  are solutions,, this rank is the codimension of the affine space of
  solutions.

So let's proceed  to row reduction of the augmented matrix:
\begin{align}
\begin{bmatrix}
1&-\lambda&1\\
\lambda-1&-2&1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&-\lambda&1\\
0&\lambda^2-\lambda-2&2-\lambda
\end{bmatrix}
\end{align}
The polynomial $\lambda^2-\lambda-2$ has $-1$ and $2$ as roots. So:


*

*If $\lambda\ne -1,2$, there is exactly one solution.

*If $\lambda= -1$, the transformed augmented matrix is $\;\begin{bmatrix}
1&1&1\\
0&0&3
\end{bmatrix}$, so there is no solution.

*If $\lambda= 2$, it is $\;\begin{bmatrix}
1&-2&1\\
0&0&0
\end{bmatrix}$, so the set of solutions is an affine line.


.
