Determinants of matrices over the finite field $\Bbb{Z}_q$

Question

Let $\Bbb{Z}_q$ be a finite field, where $q$ is prime number. Let

• $A_1$ be a set of $n \times n$ matrices such that $\det(M) = 1$, for any matrix $M \in A_1$

• $A_2$ be set of $n \times n$ matrices such that $\det(M) = 2$, for any matrix $M \in A_2$

and so forth. Is $|A_1| = |A_2|= \cdots =|A_{q-1}|$? Need proof or counterexample. I know that $|A_0|<>|A_1| $

Answer 1

Hint: Let $b\in\Bbb{Z_q}, b\neq0$. Let $X(b)$ be an $n\times n$ matrix with entry $b$ at position $(1,1)$, the other diagonal elements equal to $1$, all the non-diagonal elements equal to $0$. Show that

• $\det X(b)=b$
• The mapping $Y\mapsto YX(b)$ is a bijection from the set $A_1$ to the set $A_b$.