TITLE: Proving $\det A = 1$
QUESTION [6 upvotes]: Given a real invertible $2 \times 2$ matrix $A$ with $A + A^{-1} = I$, I need to prove that $\det A = 1$.
I know how to prove that $\det A = \frac{1}{\det A^{-1}}$, but don't have access to the fact that the determinant of a sum is the sum of the determinants (but only multiplicativity). Is there another way to prove this?

REPLY [8 votes]: Multiply by $A$:
$$
A^2-A+I=0
$$
Multiply by $A+I$:
$$
A^3+I=0
$$
Since $A^3=-I$, we get
$$
\det(A)^3=\det(-I)=1
$$
since the matrix is $n\times n$ for $n$ even.
$A$ is real, so $\det(A)$ is real; that is,
$$
\det(A)=1
$$