invariant lines of affine transformation I will write a test in linear algebra soon, this is my preparation, please check if my solution is correct.

Find all invariant lines of affine transformation $A\colon\mathbb{R}^2\to\mathbb{R}^2$ given by $A(x,y)=(2x+y-1,y+1)$.

My solution:
Let $B$ be the linear part of $A$, i.e. $B(x,y)=(2x+y,y)$.
Then $B$ has two eigenvalues: $1$ and $2$, with eigenvectors $(1,-1)$ and $(1,0)$ respectively.
For the first eigenvector:
I have to find a point $(a,b)$ such that the line $\ell:\ (a,b)+t(1,-1)$ is invariant. The condition $A(\ell)\subset\ell$ gives me $a=-1$, $b=1$ (after some computations), so $(-1,1)+t(1,-1)$ is an invariant line.
For the second eigenvector:
I have to find a point $(a,b)$ such that the line $\ell:\ (a,b)+t(1,0)$ is invariant. 
The condition $A(\ell)\subset\ell$ gives me a contradiction ($b+1=b$ appears there).
Hence $(-1,1)+t(1,-1)$ is the only invariant line of $A$.
 A: Looks good to me, although you could simplify the invariant line to $t(1,-1)$ since it passes through the origin.  
For a somewhat different approach to this problem, one can pass to homogeneous coordinates on the projective plane. The homogeneous matrix of the transformation is $$M=\begin{bmatrix}2&1&-1\\0&1&1\\0&0&1\end{bmatrix}.$$ Lines are covariant, i.e., if points transform as $\mathbf p'=M\mathbf p$, then lines transform as $\mathbf l'=M^{-T}\mathbf l$, so we’re looking for eigenvectors of $M^{-T}$. This matrix is pretty easy to invert, giving $$M^{-T}=\begin{bmatrix}\frac12&0&0\\-\frac12&1&0\\1&-1&1\end{bmatrix}$$ with eigenvalues $1$ and $\frac12$ (consistent with the eigenvalues that you found for the linear part of $M$).  
For $\lambda=1$, we can see by inspection that $[0,0,1]^T$ is an eigenvector, but this is the line at infinity, which is fixed by any affine transformation of the plane. We also have $$M^{-T}-I = \begin{bmatrix}-\frac12&0&0\\-\frac12&0&0\\1&-1&0\end{bmatrix},$$ which we can see at a glance has rank 2, so there are no other linearly independent eigenvectors of $1$.  
Turning to the other eigenvalue, you might be able to spot that the sum of the first two columns is $\left[\frac12,\frac12,0\right]^T$, so $[1,1,0]^T$ is an eigenvector. (You can of course find this eigenvector in the more usual way, too.) This corresponds to the line $x+y=0$, or, in parametric form, $t(1,-1)^T$.
