Question 1183382
The LU decomposition of B states that B = LU where L is a lower triangular matrix and U is an upper triangular matrix.  
An even better decomposition is the LU decomposition with partial pivoting, which says that 

{{{PB = LU}}}

for some permutation matrix P.  One property of permutation matrix is that it is always orthogonal, i.e., {{{P^T*P = P*P^T  = I}}}.  
In other words, 

{{{P^(-1) = P^T}}}, or its transpose is its own inverse.

Hence, {{{B = P^(-1)*LU = P^T*LU}}}

Note that by symmetry, {{{P^T}}} is also orthogonal.

Now none of the diagonal elements of U is equal to 0, since the determinant of B is 1. This implies that U can be decomposed into

{{{U = (diag(U))*U[1]}}}, 

where diag(U) is a diagonal matrix whose diagonal elements and their locations are the same as those of U, 
and {{{U[1]}}} is an upper triangular matrix where all diagonal elements are  1.

Now {{{QL = U[2]}}}, or {{{L = Q^T*U[2]}}} where Q is a permutation matrix that turns L into an upper triangular matrix {{{U[2]}}}.

==>  {{{B = P^T*Q^T*U[2]*diag(U)*U[1] = (QP)^T*U[2]*diag(U)*U[1]}}}.

Similarly, {{{U[2] = diag(U[2])*U[3]}}}, where {{{diag(U[2])}}} is a diagonal matrix whose diagonal elements and their locations are the same as those of {{{U[2]}}}, 
and {{{U[3]}}} is an upper triangular matrix where all diagonal elements are 1.

==> {{{B = (QP)^T*diag(U[2])*U[3]*diag(U)*U[1]

= (QP)^T*diag(U[2])*(U[3]*diag(U))*U[1] 

= (QP)^T*diag(U[2])*(diag(U)*U[4])*U[1]}}}.

where {{{U[4]}}} is still an upper triangular matrix with diagonal elements equal to 1.

==> {{{B = (QP)^T*(diag(U[2])*diag(U))*(U[4]*U[1])}}}.

Now {{{det(QP)^T=1}}}, because its rows are just rearrangements of the rows of I; 
{{{U[4]*U[1]}}} is an upper triangular matrix whose diagonal elements are all 1, hence its determinant is equal to 1; 
and finally {{{diag(U[2])*diag(U)}}} has the same determinant as LU, and therefore equal to 1.

Hence, let {{{K = (QP)^T}}}, {{{A = diag(U[2])*diag(U)}}}, and {{{N = U[4]*U[1]}}}, and the statement is proved.