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)

• Any triangular matrix with ones along the diagonal has unit determinant. – amd Oct 26 '17 at 5:37

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.

• Ok, I think i understand it now. The analogy helped. Also side question if you don't mind me asking does the transpose of a orthogonal matrix = the inverse? – LKRC Oct 26 '17 at 4:30
• @LKRC: Yes, that's the definition of what it means to be an orthogonal matrix. – Clive Newstead Oct 26 '17 at 4:53

Try the matrix $$\begin{bmatrix}1&1\\0&1\end{bmatrix}$$ with determinant $1$. Are the columns orthogonal?

• .............,,no? – LKRC Oct 26 '17 at 4:24
• @LKRC no they are not, their dot product is 1. – qbert Oct 26 '17 at 4:29

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)