This article - https://doi.org/10.1016/0370-2693(88)91216-6 - states that for a $3 \times 3$ unitary matrix $A$, we have $|A_{33}| = |(A_{11}A_{22}-A_{12} A_{21})|$, where $A_{ij}$ stands for the element in row $i$ and column $j$ of matrix $A$ and |.| stands for the modulus.
Why is this true?
I tried expanding the square of the right-hand side of the equality but I got a term with $\Re(A_{11}A_{22}A_{12}^*A_{21}^*)$ that I don't think can easily be related to $|A_{33}|$ so this may not be the correct procedure.
Note the formula $A^{-1} = \frac{\mathrm{adj}(A)}{\det(A)}$, where $\mathrm{adj}(A)$ is the adjugate matrix of $A$. Then, $$\begin{pmatrix}* &* &* \\ * &* &* \\ * &* & \overline{A_{33}}\end{pmatrix} = A^* = A^{-1} = \frac{1}{\det{A}}\begin{pmatrix}* &* &* \\ * &* &* \\ * &* &(A_{11}A_{22}-A_{12}A_{21}) \end{pmatrix},$$ where $*$ represents some matrix entries we don't care about. Comparing the bottom right corners, we get $\overline{A_{33}} = \frac{A_{11}A_{22}-A_{12}A_{21}}{\det(A)}$. Taking the magnitude of both sides gives $|A_{33}| = \frac{|A_{11}A_{22}-A_{12}A_{21}|}{|\det(A)|}$, but $1=\det(AA^*)=|\det(A)|^2$, so this simplifies to the desired result.