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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

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  • $\begingroup$ What do you know about determinants? Do you know that $\det(A^T) = \det(A)$? And that $\det(A^{-1})\det(A) = 1$? $\endgroup$ – bames Feb 21 '18 at 7:44
  • $\begingroup$ 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 $\endgroup$ – mrnobody Feb 21 '18 at 7:44
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    $\begingroup$ @mrnobody That's not such a useful approach if the matrix is $3\times 3$ or bigger. $\endgroup$ – Lord Shark the Unknown Feb 21 '18 at 7:47
  • $\begingroup$ always try to include your attempts and thoughts when you post a question if possible. and yup, mathjax is awesome. $\endgroup$ – Siong Thye Goh Feb 21 '18 at 7:52
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Hint:

$$\det(AB)=\det(A)\det(B)$$

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Ingredients:

  • What is the definition of inverse? Write it down.
  • $\det(AB)=\det(A)\det(B)$
  • $\det(A)=\det(A^T)$
  • $\det(I)=1$
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    $\begingroup$ GOT ITTT thanks $\endgroup$ – mrnobody Feb 21 '18 at 7:53

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