Minors of orthonormal matrices are easy to compute Here is an intriguing fact:
Take any $3\times 3$ orthogonal matrix $A$. For example, take
$$
A=\begin{pmatrix}
1/\sqrt{14} & \sqrt{2/7} & 3/\sqrt{14} \\ 
2 \sqrt{13/105} & -5 \sqrt{5/273} & 8/\sqrt{1365}  \\
\sqrt{13/30} & \sqrt{10/39} & -11/\sqrt{390}
\end{pmatrix}
$$
which has been produced by applying the Gram–Schmidt process to some random choice of vectors.
Compute the determinant of any $2\times 2$ submatrix. For example, the one consisting of the four corners:
$$
\begin{vmatrix}
1/\sqrt{14}& 3/\sqrt{14} \\
\sqrt{13/30} & -11/\sqrt{390}
\end{vmatrix}
=
-5 \sqrt{5/273}
$$
Note that the result is the middle element of $A$.
Doing this for any submatrix of any $A$, we always get $\pm$ the entry that completes the submatrix. Why is that?
 A: Given a square matrix $M \in SO_n,$ so $\det M = 1$ and $MM^T = M^T M = I,$ decomposed as illustrated  with square blocks $A,D$ and rectangular blocks $B,C,$
$$M  = \left( \begin{array}{cc} 
A & B \\\  
 C & D 
 \end{array} \right) ,$$
then   $\det A = \det D.$
What this says is that, in Riemannian geometry with an orientable manifold, the Hodge star operator is an isometry, a fact that has relevance for Poincare duality.
http://en.wikipedia.org/wiki/Hodge_duality
http://en.wikipedia.org/wiki/Poincar%C3%A9_duality
But the proof is a single line:
$$ \left( \begin{array}{cc}  A & B \\\  0 & I  \end{array} \right) 
\left( \begin{array}{cc}  A^t & C^t \\\ B^t & D^t  \end{array} \right)   = 
\left( \begin{array}{cc}  I & 0 \\\  B^t & D^t  \end{array} \right). $$
Let's see; given any submatrix defined by a choice of $p$ rows and $p$ columns, we can make a "dual" submatrix of the other  $n-p$ rows and $n-p$ columns. By permuting the row and permuting the columns, we can make the $p$ by $p$ matrix be the upper left square corner and the other matrix the lower right. I suppose some more care is needed about $\pm$ signs; so far this just shows that the determinants of the originals have the same absolute value. EDIT: this is as it should be. If we take an off-diagonal element in the matrix of the question, for instance, the 1,2 position with $\sqrt{2/7},$ the determinant of the submatrix with positions 2,1; 2,3; 3,1; 3,3 is negative. More generally, we can make some row transpositions and some column transpositions, always one of the $p$ switched with one of the $n-p$ so as to keep the submatrices intact, and get the $p$ by $p$ as the upper left corner, having not changed the determinants. However, if the total number of transpositions was odd, we end up with $\det M = -1$ and $\det A = - \det D.$ 
A: The inverse of a matrix is given by:
$$
(A^{-1})_{ij}=\frac{1}{\det A}A'_{ji}
$$
where $A'_{ji}$ is the signed minor of the element $a_{ji}.$
Now, for an orthogonal matrix, we have
$$
Q'_{ji}=\det Q (Q^{-1})_{ij}=\pm (Q^T)_{ij}=\pm Q_{ji}.
$$
