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This concerns $RQ$ decompositions of square matrices.
Suppose you have a $n \times n$ matrix $A$ and want to compute the RQ decomposition. If you know how to compute QR decompositions, you will just need a few transpositions / row-column permutations.
Note that given the matrix $P := \begin{bmatrix}
& & 1\\
& \iddots & \\
1 & & \\
\end{bmatrix}$ (this is different to the camera matrix mentioned in the question), we get that $AP$ reverses the order of columns of $A$ and $PA$ reverses the order of rows. Also note that $P^T = P$ and $PP = E_n$, so $P^{-1} = P = P^T$, in particular $P$ is orthogonal.
Consider the following algorithm:
i.) Compute $\tilde A := PA$ (i.e. reverse rows of $A$)
ii.) Compute decomposition of $\tilde A ^T = \tilde Q \tilde R$
iii.) Set $Q := P \tilde Q^T$ (i.e. reverse rows of $\tilde Q^T$, note that $Q$ is orthogonal)
iv.) Set $R := P \tilde R^T P$
In step iv.) the following happens: $\tilde R$ is an upper triangular matrix. By transposing it, it becomes a lower triangular matrix. So we reverse rows and columns and obtain again an upper triangular matrix $R$. See sketch (start with lower triangular, reverse rows, then revere columns).
$$\begin{bmatrix}
* & \cdot & \cdot \\
* & * & \cdot \\ * & * & *
\end{bmatrix} \to
\begin{bmatrix}
* & * & * \\
* & * & \cdot \\
* & \cdot & \cdot
\end{bmatrix} \to
\begin{bmatrix}
* & * & * \\
\cdot & * & * \\
\cdot & \cdot & * \\
\end{bmatrix}
$$
Altogether $R$ and $Q$ yields our decomposition:
$$RQ = (P \tilde R^T P)(P \tilde Q^T) = P \tilde R^T \tilde Q^T = P(\tilde Q \tilde R)^T = P(\tilde A ^T)^T = P \tilde A = PPA = A$$