= Solution
Write $J_1,J_2,J_3$ for the actual successive <rotations> $J(1,2)$, $J(1,3)$, and $J(2,3)$, each selected using the currently transformed <matrix>. The first eliminates position $(2,1)$; the second eliminates $(3,1)$ while leaving row two unchanged. After these, the first column is $(r_{11},0,0)^T$. The final <rotation> of rows two and three preserves both zeros in that column and eliminates $(3,2)$. Hence
$$
\boxed{R=J_3J_2J_1A\text{ is upper triangular}.}
$$
A lower-triangular starting <matrix> does not <force> any further zeros above the diagonal. For example, the lower-triangular <matrix> with all entries on and below its diagonal equal to one gives, with the nonnegative <pivot> convention,
$$
R=\begin{pmatrix}\sqrt3&2/\sqrt3&1/\sqrt3\\0&\sqrt{2/3}&1/\sqrt6\\0&0&1/\sqrt2\end{pmatrix},
$$
which has all upper-triangular entries nonzero.
Inverting the product of <rotations> yields the <QR factorization>
$$
\boxed{Q=J_1^TJ_2^TJ_3^T,\qquad A=QR.}
$$
The order reverses when taking the <transpose>. If $A$ is <orthogonal>, then $R$ is both <orthogonal> and <upper triangular>. Its first column forces $|r_{11}|=1$ and <orthogonality> with the other columns forces $r_{12}=r_{13}=0$. Applying the same argument to the remaining block makes $R$ diagonal with entries of modulus one.
Our elimination rule makes $r_{11}$ and $r_{22}$ positive when nonzero, so they are both one. Every <rotation> has <determinant> one; thus $\det R=\det A$ and
$$
\boxed{R=\operatorname{diag}(1,1,\det A).}
$$
Without a fixed pivot-sign convention one could only specify a diagonal sign <matrix>. The convention chosen in part (c) determines the precise signs above.
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