Let $A\in \mathcal {M}_n(\mathbb{F}_2)$ such that $A^3=I$. Find $\det(A)$ a) Let $A\in \mathcal {M}_n(\mathbb{R})$ such that $A^3=I$. Find $\det(A)$
b) Let $A\in \mathcal {M}_n(\mathbb{C})$ such that $A^3=I$. Find $\det(A)$
c) Let $A\in \mathcal {M}_n(\mathbb{F}_2)$ such that $A^3=I$ find $\det(A)$
My work:
 As $A^3=I$ then $A=I$.
Then:
a) $\det(A)=1$
b) $\det(A)=1$
c) $\det(A)=?$
I don't know how calculate $\det(A)$ in the third case. Can someone help me?
 A: It is possible to have $A^3 = I$ and $A \neq I$ at the same time. For instance, if your field is $\mathbb{C}$, then the diagonal matrix with $e^{2 \pi i /3}$ on the diagonal will have this property. In relation to this example, also note that over the complex numbers $(\det A)^3 = 1$ does not imply that A has determinant 1, but rather that the determinant is any 3rd root of unity. 
Now, regarding $\mathbb{F}_2$: Over this field, the equation $x^3=1$ only has one solution. It's not 0, so it has to be$\ldots$
A: The assertion $A^3=I$ then $A=I$ is not true in general. 
For instance in $\mathcal{M}_n(\mathbb{C})$, consider $A=\omega I$ where $\omega=e^{\frac{2i\pi}{3}}$ is such that $\omega^3=1$.
Using the multiplicative property of the determinant one get $1=det(I)=det(A^3)=det(A)^3$ so in each case the determinant of $A$ is a number $a$ such that $a^3=1$. In each case it gives :
a) $a^3=1$ so $a=1$ 
b) $a^3=1$ so $a\in\{1,\omega,\omega^2\}$ where $\omega$ is defined above
c) $a^3=1$ so $a=1$
For each case there is a different reason leading to the conclusion. 
a) the cubical root is the bijection from $\mathbb{R}$ to $\mathbb{R}$ so there is exactly $1$ solution $a$ to $a^3=1$, which is obviously $1$.
b) A polynomial of degree $n$ (so non zero) over a field has at most $n$ roots. $\mathbb{C}$ is a field and I gave you explicitly $3$ roots.
c) $\mathbb{F}_2$ has two elements and it's not hard to try them both. As $0^3=0$ and $1^3=1$ we are done. 
