Suppose that $P$ is an $n \times n$ matrix such that $P^T(P^3)=I$ , where $I$ is the $n \times n$ identity matrix. Find all possible values of $\det(P)$.
I solved to $P^T=P^{-1}(P^{-1})(P^{-1})$ where $\det(P)=\det(P^{-1}(P^{-1})(P^{-1}))$ but I'm pretty sure it's incorrect.
On
Edit: I just noticed that you probably meant the 'T' to denote the transpose of $P$. In that case the train of thought is similar, but now along with the property $\det AB=\det A\det B$ you need also to use that $\det A^t=\det A$. This being said the calculations are easy: $$ 1=\det I=\det (P^t)\det (P^3)=\det P(\det P)^3=(\det P)^4\Rightarrow \det(P)\in\{1,i,-1,-i\}. $$
You don't need to cancel $P^3$ (as an aside; notice that that would need a justification that $P$ is not singular). Just take determinants in $P^TP^3=I$ (another aside; notice that $P^TP^3=P^{T+3}$) to arrive at $$ 1=\det I=\det (P^{T+3})=\det(P)^{T+3}. $$ So $\det{P}$ satisfies the equation $1=x^{T+3}$; i.e. $\det(P)$ is a $(T+3)-$root of unity. More explicitly $$ \det(P)=e^{2\pi i\frac{k}{T+3}} $$ for some $0\leq k<T+3$.
$\det (P^t P^3)=1\implies \det P^t\det P^3=1\implies (\det P)^4=1\implies \det P=1\text {or}\det P=-1$