On page 392 of Computer vision by Simon Prince (http://web4.cs.ucl.ac.uk/staff/s.prince/book/book.pdf), equation 15.2 has the following expression for the Euclidean transformation in homogeneous coordinates:
$$ \lambda \begin{bmatrix} x\\y\\1 \end{bmatrix} = \begin{bmatrix} \phi_x&\gamma&\delta_x\\ 0 &\phi_y&\delta_y\\ 0&0&1 \end{bmatrix} \begin{bmatrix} w_{11}&w_{12}&\tau_x\\ w_{21}&w_{22}&\tau_y\\ 0&0&D \end{bmatrix} \begin{bmatrix} u\\v\\1\end{bmatrix} $$
In what follows, he switches the position of $1$ and $D$:
$$ \lambda \begin{bmatrix} x\\y\\1 \end{bmatrix} = \begin{bmatrix} \phi_x&\gamma&\delta_x\\ 0 &\phi_y&\delta_y\\ 0&0&D \end{bmatrix} \begin{bmatrix} w_{11}&w_{12}&\tau_x\\ w_{21}&w_{22}&\tau_y\\ 0&0&1 \end{bmatrix} \begin{bmatrix} u\\v\\1\end{bmatrix} $$
with the comment that the last equation is unchanged.
My question is this: the other two equations do seem to change when you make this switch, so firstly, why can we make this switch? Secondly, is there a geometric explanation underlying this switch?
Edit:
Here are the multiplied out matrices:
$$ \begin{bmatrix} \phi_x&\gamma&\delta_x\\ 0 &\phi_y&\delta_y\\ 0&0&1 \end{bmatrix} \begin{bmatrix} w_{11}&w_{12}&\tau_x\\ w_{21}&w_{22}&\tau_y\\ 0&0&D \end{bmatrix} = \begin{bmatrix} \phi_x w_{11} + \gamma w_{21}&\phi_x w_{12} + \gamma w_{22}&\phi_x \tau_x + \gamma \tau_y + D\delta_x\\ \phi_y w_{21}& \phi_y w_{22}&\phi_y \tau_y + D\delta_y\\ 0&0&D \end{bmatrix} $$
$$ \begin{bmatrix} \phi_x&\gamma&\delta_x\\ 0 &\phi_y&\delta_y\\ 0&0&D \end{bmatrix} \begin{bmatrix} w_{11}&w_{12}&\tau_x\\ w_{21}&w_{22}&\tau_y\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} \phi_x w_{11} + \gamma w_{21}&\phi_x w_{12} + \gamma w_{22}&\phi_x \tau_x + \gamma \tau_y + \delta_x\\ \phi_y w_{21}& \phi_y w_{22}&\phi_y \tau_y + \delta_y\\ 0&0&D \end{bmatrix} $$
As can be seen in the last column, the first two equations have changed.
Edit 2:
I have checked with my lecturer and he thinks that there might be an error in the book.