Linear Algebra: Proof of determinants I need to show that if a and b are two non-zero members of Field q, then the number of nxn matrices with determinant a is the same as the number with determinant b.  
Im really not sure how how to approach this proof, please help. 
 A: Multiplying the first row of a matrix by $ab^{-1}$ is a bijection between the set of matrices of determinant $b$ and the set of matrices of determinant $a$.  
Added
Let $S,T$ be the set of matrices of determinant $b$ respectively $a$.
Let $f :S \to T$ be defines by
$$f(  \begin{bmatrix}
a_{11}&a_{12}&...&a_{1n}\\
a_{21}&a_{22}&...&a_{2n}\\
...&....&...&.....\\
a_{n1}&a_{n2}&...&a_{nn}\\
\end{bmatrix})= \begin{bmatrix}
ab^{-1}a_{11}&ab^{-1}a_{12}&...&ab^{-1}a_{1n}\\
a_{21}&a_{22}&...&a_{2n}\\
...&....&...&.....\\
a_{n1}&a_{n2}&...&a_{nn}\\
\end{bmatrix}$$
Then, by dividing the first row of $f(A)$ by $ab^{-1}$ you get that
$$\det(f(A))=ab^{-1}\det(A) \,;\, \forall A \in S \,.$$
From here you should get imediatelly that $f$ is well defined and bijection.
Or you can also construct its inverse, exactly the same way.....
A: Try thinking about the homomorphism: $$\phi : GL_n(\mathbb{F})\rightarrow \mathbb{F}\setminus{\{0\}}$$ $$\phi(A)=\det (A)$$ What is the kernel of $\phi$?  What are the cosets of the kernel?  What do elements of the same coset have in common?
