Showing homomorphism for $\theta: GL_2 (\Bbb Q) \rightarrow \Bbb Q\setminus \{0\}$ given by $\theta(A) = \det A$. 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?)
$\ker \theta = A: \det A =1$
 A: Start working from the other end. It’s usually better either to work from the more complicated end or to work on both ends of the calculation simultaneously.
You know that $$AB=\pmatrix{a_1&a_2\\a_3&a_4}\pmatrix{b_1&b_2\\b_3&b_4}=\pmatrix{a_1b_1+a_2b_3&a_1b_2+a_2b_4\\a_3b_1+a_4b_3&a_3b_2+a_4b_4}\;,$$
so 
$$\begin{align*}
\det AB&=(a_1b_1+a_2b_3)(a_3b_2+a_4b_4)-(a_1b_2+a_2b_4)(a_3b_1+a_4b_3)\\
&=\color{red}{a_1b_1a_3b_2}+a_1b_1a_4b_4+a_2b_3a_3b_2+\color{blue}{a_2b_3a_4b_4}\\
&\qquad-\color{red}{a_1b_2a_3b_1}-a_1b_2a_4b_3-a_2b_4a_3b_1-\color{blue}{a_2b_4a_4b_3}\\
&=a_1b_1a_4b_4+a_2b_3a_3b_2-a_1b_2a_4b_3-a_2b_4a_3b_1\\
&=a_1a_4b_1b_4-a_1a_4b_2b_3+a_2a_3b_2b_3-a_2a_3b_1b_4\\
&=a_1a_4(b_1b_4-b_2b_3)-a_2a_3(b_1b_4-b_2b_3)\\
&=(a_1a_4-a_2a_3)(b_1b_4-b_2b_3)\\
&=\det A\det B\;.
\end{align*}$$
A: You haven't multiplied out $AB$ ---- you have to do that, then you can compute $\det(AB)$ and see whether it equals $\det A\det B$. 
A: $$AB = \begin{pmatrix}
        a_1b_1+a_2b_3 & a_1b_2+a_2b_4 \\
        a_3b_1+a_4b_3 & a_3b_2+a_4b_4 \\
        \end{pmatrix}$$
So $$\det AB = (a_1b_1+a_2b_3)(a_3b_2+a_4b_4) - (a_1b_2 + a_2b_4)(a_3b_1+a_4b_3)$$
$$= (a_1a_3b_1b_2 + a_1a_4b_1b_4 + a_2a_3b_2b_3 + a_2a_4b_3b_4) - (a_1a_3b_1b_2+a_1a_4b_2b_3+ a_2a_3b_1b_4+ a_2a_4b_3b_4)$$
$$= a_1a_4b_1b_4+a_2a_3b_2b_3-a_1a_4b_2b_3-a_2a_3b_1b_4 $$
$$= a_1a_4(b_1b_4-b_2b_3) -a_2a_3(b_1b_4-b_2b_3) $$
$$= (a_1a_4-a_2a_3)(b_1b_4-b_2b_3) $$
$$= \det A \det B.$$
