Proving $\det A = 1$ 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?
 A: 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,
$$
\det(A)=1
$$
A: Multiply both sides of $A+A^{-1}=I$ by $A$ to get $A^2+I=A$ or $A^2-A+I=0$, and note that,
for an $n\times n$ matrix, the constant coefficient of the characteristic polynomial is $(-1)^n\det(A)$.
Addendum in response to comment:
Here is another way.
Let $A=\pmatrix{a&b\\c&d}$.  Then $A^{-1}=\dfrac{\pmatrix{d&-b\\-c&a}}{\det A}.$
$A+A^{-1}=I=\pmatrix{1&0\\0&1}\implies$
$a+\dfrac d{\det A}=1=d+\dfrac a{\det A}$ and $ b-\dfrac b{\det A}=0=c-\dfrac c{\det A}$
$\implies \det A=1$ or $a-d=b=c=0$, but we can't have the latter,
since then $\pmatrix{a&0\\0&a}+\pmatrix{a&0\\0&a}^{-1}=I$ would imply $a+\dfrac1a=1$, which has no real solutions.
A: Post multiplying the equation on both sides by $A$ we get
$$A^2 -A +I=0$$
Therefore an associated annihilating polynomial is
$$p(\lambda)=\lambda^2 -\lambda +1$$
Observing that it has degree $2$ and is irreducible over $\mathbb{R}$ it follows that it's the characteristic polynomial.
Clearly the product of eigenvalues of $A =(-\omega)(-\omega^2)=1=det(A)$
where $\omega$ denotes a third root of unity
Edit:I really appreciate @Christoph's criticism and help in finding and ultimately trying to fill gaps.
There's another slightly different (almost made up) way, which is by noting that the characteristic polynomial associated to $A$ and $A^{-1}$ is identical which means $det(A)=det(A^{-1})=1/det(A)$ which implies we've the possibilities $det(A)=\pm1$ , we can rule out the $-1$ by looking at the coefficients of the characteristic polynomial
