# Prove that if $M$ is orthogonal, then $\det(M)= \pm 1$

Recall that a matrix $M$ is orthogonal if it is square and $M^TM =I$. Prove that $\det(M) = \pm 1$ for every orthogonal matrix $M$.

Not sure how to go about showing this for every orthogonal matrix

• Hint: What do you know about the determinant of $M^T$ for an arbitrary square matrix $M$? – amd Apr 3 '18 at 4:27
• Hint: $\det M^T = \det M$. – Henning Makholm Apr 3 '18 at 4:27
• I edited your post to patch up the $\LaTeX$ a tad. Cheers! – Robert Lewis Apr 3 '18 at 4:38

Well, we know that

$\det M^T = \det M, \tag 1$

and we are given that

$M^T M = I, \tag 2$

so

$\det M^TM = \det I = 1; \tag 3$

also,

$\det M^TM = \det M^T \det M, \tag 4$so using (1), (3) and (4):

$(\det M)^2 = \det M \det M = \det M^T \det M =\det M^TM = 1, \tag 5$

and so we must have

$\det M = \pm 1. \tag 6$