relationship between det and orthogonality I know that all orthogonal matrices have a $\det$ of either $1$ or $-1$, but I need to know if this statement goes both ways. 
My question:  If $\det$ of a matrix $= -1$ or $1$ does that matrix HAVE TO BE orthogonal? and if not, why?(and if $\det = -1$ or $1$ and not orthogonal is possible then  can I get an example of one such matrix)
 A: As intuition behind determinant is about volume of the shape with sides of the vectors, so, we can have a shape with volume equal to one, although its sides are not orthogonal.
A: Try the matrix 
$$
\begin{bmatrix}1&1\\0&1\end{bmatrix}
$$
with determinant $1$. Are the columns orthogonal?
A: In an orthogonal matrix every column (and row) should be a unit vector.
So no entry in a column (or row) can exceed  $+1$
Using  this remark we can easily generate lot of matrices of determinant 1 that are not orthogonal
Take two consecutive integers that are not prime and factorize them both: for example $44,45$ factorized as $44= 11\times 4,\ 45= 5\times 9$. So $1=45-44 = 5\times9-11\times4$.
The  matrix $A=\pmatrix{5& 11\cr 4 & 9\cr}$ has determinat 1 and entries are bigger than 1 so its columns are not unit vectors.
In general take an $n\times n$ matrix $B$ whose determinant $\Delta$ is a positive number.
Now divide each entry of the first row by $\Delta$. This new matrix will have determinant 1 and most of the times non-orthogonal (arrange to have 2 as an entry in the 2nd row, for example)
A: I wish it did go both ways but the determinant mixes vector magnitude information with vector direction information.
Orthogonality is a relationship between vector directions but completely ignores the magnitudes of the vectors. If you can assume vector magnitudes are all one (i.e. all unit vectors), you can use the determinant to measure degree of orthogonality between vectors.
