Show that this map is a group homomorphism and find its kernel:
$\theta: GL_2 (\Bbb Q) \rightarrow \Bbb Q\setminus \{0\}$ given by $\theta(A) = \det A.$
My attempt:
Let $A = \begin{pmatrix} a_1 & a_2 \\ a_3 & a_4 \\ \end{pmatrix}$ then $ \theta (A) =\det A = a_1a_4 - a_2a_3$ And let $B \in GL_2 (\Bbb Q)$ such that B = \begin{pmatrix} b_1 & b_2 \\ b_3 & b_4 \\ \end{pmatrix} and $\theta(B) = \det B = b_1b_4 - b_2b_3$
Then checking for homomorphism...
$ \begin{align} \theta(A)\theta(B)= \det A \det B & = \ (a_1a_2-a_3a_4)(b_1b_4 - b_2b_3) \\ & = a_1a_2b_1b_2 - a_3a_4b_1b_4 - a_1a_2b_2b_3 + a_3a_4b_3b_4\\ & = \det(AB) = \theta(AB) \end{align}$
(to be honest I couldn't actually figure out how $\det A\det B$ became $\det AB$ with the method I used. i.e. the expansions were just not working out. Is there a better way of doing this? And am I horrificaly wrong?)
$\ker \theta = A: \det A =1$