Epipolar geometry - Fundamental matrix derivation (Hartley, Zisserman) I have a question to the following derivation of the fundamental matrix by Hartley and Zisserman in "Multiple View Geometry in computer vision" (Link, page 5):

Why is it possible to do the very last transformation step? What's the trick here?
 A: The last step employs the identity $(\det{\mathbf{A}})\,[\mathbf{z}]_\times = \mathbf{A}^\top[\mathbf{A}\mathbf{z}]_\times\mathbf{A}$, which holds for all $\mathbf{z}\in\mathbb{R}^3$ and $\mathbf{A}\in\mathbb{R}^{3\times 3}$.
(There seems to be a scalar factor $\det{\mathbf{K}}$ missing, presumably intentionally left out as everything is only up to scale anyway, and $\det{\mathbf{K}}\neq 0$ for the intrinsic calibration matrix $\mathbf{K}$.)

The identity can be derived in the following way, using the fact that the determinant of a $3\times 3$ matrix equals the triple product of its columns:
Let $\mathbf{A},\mathbf{B}\in\mathbb{R}^{3\times 3}$, and let $\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3$ be the columns of $\mathbf{B}$.
Now, on the one hand,
$$
\det(\mathbf{A}\mathbf{B})
= \det{\mathbf{A}}\cdot\det{\mathbf{B}}
= (\det{\mathbf{A}}) \cdot \mathbf{b}_1^\top(\mathbf{b}_2 \times \mathbf{b}_3)
= \mathbf{b}_1^\top \big((\det{\mathbf{A}})\,[\mathbf{b}_2]_\times\big) \mathbf{b}_3.
$$
On the other hand,
$$
\det(\mathbf{A}\mathbf{B}) =
\begin{vmatrix}
    \mathbf{A}\mathbf{b}_1 & \mathbf{A}\mathbf{b}_2 & \mathbf{A}\mathbf{b}_3
\end{vmatrix}
= (\mathbf{A}\mathbf{b}_1)^\top \big((\mathbf{A}\mathbf{b}_2) \times (\mathbf{A}\mathbf{b}_3)\big)
= \mathbf{b}_1^\top \big(\mathbf{A}^\top [\mathbf{A}\mathbf{b}_2]_\times \mathbf{A}\big) \mathbf{b}_3.
$$
This holds for any $\mathbf{B}$, so $(\det{\mathbf{A}})\,[\mathbf{b}_2]_\times = \mathbf{A}^\top [\mathbf{A}\mathbf{b}_2]_\times \mathbf{A}$.
Finally, rename $\mathbf{b}_2$ to something nicer, like $\mathbf{z}$.
