Question about proof that $\det(Q)=\pm 1$ for orthogonal matrix $Q$

Question

For orthogonal matrix, $Q$, $QQ^T=I\tag{1}$ Wikipedia states that $1=\det(I)=\det\left(Q^\mathrm{T}Q\right)=\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2$

But I can't understand how the last equality (on the right) follows: $\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2$

Is it the case that $\det\left(Q^T\right)=\det\left(Q\right)$? This would imply $Q^T=Q$, but taking the inverse of $(1)$ implies that $Q^T=Q^{-1}\ne Q$ which is a contradiction.

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Before asking this question I searched this site and read this strongly related question. In one of the comments and the answer it is stated right from the start that $\det M^T=\det M$, but I still don't understand how this can be true.

Answer 1

Yes, for any square matrix $A\in M_n(K)$, it is true that $\operatorname{det} A = \operatorname{det} A^T$, and this doesn’t imply that $A=A^T$. In general, two matrices can be distinct and yet have the same determinant, for example: \begin{gather*} A = \begin{pmatrix} 1 & 3 \\ 1 & 2 \end{pmatrix} \\ B = \begin{pmatrix} 2 & -3 \\ -3 & 4 \end{pmatrix} \end{gather*} $A$ and $B$ are distinct matrices but $\operatorname{det} A = \operatorname{det} B = -1$.

Now, if you are looking for a proof of the equality $\operatorname{det} A = \operatorname{det} A^T$ (which is widely known) take a look here

Answer 2

It's true that $\det\left(Q^T\right)=\det\left(Q\right)$ but this doesn't imply that $Q^T=Q$ (take any non-symmetric matrix) (see here for more details).