# Determinant is $1$ if the matrix has its transpose as its inverse. PROOF

We say that a square matrix $A \in \mathbb{R}^{n\times n}$ is unitary if its inverse is given by its transpose. Show that, for a unitary matrix, one has that $\det A = \pm 1$.

I would like to focus on the info given here =and not drift away into explications that are beyond my level. How, from knowing that the transpose is the inverse can we prove that $|\det A|=1$

I tried the inverse=the transpose with $a, b, c, d$ as my numbers and I tried to match up each term with their corresponding one on the other side and =then solve the equation

• What do you know about determinants? Do you know that $\det(A^T) = \det(A)$? And that $\det(A^{-1})\det(A) = 1$? Feb 21 '18 at 7:44
• so I tried the inverse=the transpose with a b c d as my numbers and I tried to match up each term with their corresponding one on the other side and =then solve the equation Feb 21 '18 at 7:44
• @mrnobody That's not such a useful approach if the matrix is $3\times 3$ or bigger. Feb 21 '18 at 7:47
• always try to include your attempts and thoughts when you post a question if possible. and yup, mathjax is awesome. Feb 21 '18 at 7:52

• $\det(AB)=\det(A)\det(B)$
• $\det(A)=\det(A^T)$
• $\det(I)=1$
$$\det(AB)=\det(A)\det(B)$$