Interpretation of a homogeneous transformation matrix of the plane I have the transformation matrix $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}$. This $3\times 3$ matrix is a homogeneous transformation matrix in $2-D$ space. 
My book says that this matrix translates the line  $y=x+1$ to $y=x$. I don't see how. Let us take the point $(a,b)$. After the translation by $-1$ along the $y$-axis, the point should become $(a,b-1)$. Now let us determine $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix} a\\b\\1\end{pmatrix}$. We get $\begin{pmatrix} a\\b\\1-b\end{pmatrix}$. After transforming this resultant matrix to 2-D form, we get $\begin{pmatrix} \frac{a}{1-b}&\frac{b}{1-b}\end{pmatrix}$. How is this equal to $(a,b-1)$?
 A: Work backwards.  The translation of $y = x - 1$ to $y = x$:
$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \end{bmatrix}$$
Affine to linear:
$$\begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1  & -1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$$
So it seems the book is representing the vectors as row vectors, $\begin{bmatrix} x & y \end{bmatrix}$, so your matrix is transposed.
A: It appears you've either put the translation elements in the wrong position or you're using column vectors instead of row vectors:
$$
\begin{pmatrix}a&b&1\end{pmatrix}
\begin{pmatrix}1&0&\color{#00A000}{0}\\0&1&\color{#00A000}{0}\\\color{#C00000}{0}&\color{#C00000}{-1}&1\end{pmatrix}
=\begin{pmatrix}a&b-1&1\end{pmatrix}
$$
or
$$
\begin{pmatrix}1&0&\color{#C00000}{0}\\0&1&\color{#C00000}{-1}\\\color{#00A000}{0}&\color{#00A000}{0}&1\end{pmatrix}
\begin{pmatrix}a\\b\\1\end{pmatrix}
=\begin{pmatrix}a\\b-1\\1\end{pmatrix}
$$
translation elements in $\color{#C00000}{\text{red}}$, perspective elements in $\color{#00A000}{\text{green}}$.
