Special orthogonal matrix uniquely determined by $n-1 \times n-1$ entries? For example, consider the specific question: Given $a_{11},a_{12},a_{21},a_{22}$ does that uniquely determine
$A=\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix}$
where $A\in SO(3)$.
 A: Hint 1: The sum of the squares of each column is $1$.
Hint 2: The third column is $\pm$ the cross product of the first two.
Hint 1 is immediately applicable to $n\times n$ matrices.
There is an $n$-dimensional analog of the cross product that extends Hint 2.
A: Note that if $A \in SO_n(\mathbb R)$ with
$$ A \; = \;  
 \left(  \begin{array}{rr}
  E & F  \\
   G  & H  
\end{array} 
  \right)  ,
$$
and submatrices $F,G$ rectangular and  $E,H$ both perfectly square, then $$ \det E = \det H. $$
I'm just saying. In your case, if you have the upper left square, it turns out that $$ a_{33} = \; a_{11} a_{22} \; - \; a_{12} a_{21}.   $$ Put another way, if you are given the upper left block and complete it, another choice for $A \in SO_n(\mathbb R)$ is to negate column $n$ and row $n,$ which means that $a_{nn}$ is negated twice and actually stays put.
Proof:
$$ \left( \begin{array}{cc}  E & F \\  0 & I  \end{array} \right) 
\left( \begin{array}{cc}  E^t & G^t \\ F^t & H^t  \end{array} \right)   = 
\left( \begin{array}{cc}  I & 0 \\  F^t & H^t  \end{array} \right). $$
Note that this is more general as both $E,H$ are allowed to have size larger than one, if $n > 3$ anyway. 
