Prove that det(AB) = det(A) det(B) in AB ∈ $GL_2(\mathbb{R} \!\,)$

Question

Prove that $\det(AB) = \det(A) \det(B) $ if $A,B \in \operatorname{GL}_2(\mathbb{R})$. Use this result to show that the binary operation in the group $\operatorname{GL}_2(\mathbb{R})$ is closed; that is, if $A $ and $ B$ are in $\operatorname{GL}_2(\mathbb{R})$, then $AB \in \operatorname{GL}_2(\mathbb{R})$.

I am totally stuck, I know these things to be true but I don't know how to show that. Any tips on how to get started?

Answer 1

Let $A=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} x & y \\ z & t \\ \end{array} \right)$.

$\det(A)=ad-bc$, $\det(B)=xt-yz$. $\det(A)\det(B)=(ad-bc)(xt-yz)=adxt-bcxt-adyz+bcyz$. $AB=\left( \begin{array}{cc} ax+bz & ay+bt \\ cx+dz & cy+dt \\ \end{array} \right)$

$\det(AB)=(ax+bz)(cy+dt)-(ay+bt)(cx+dz)=axcy+bzcy+axdt+bzdt-aycx-btcx-aydz-btdz=bzcy+axdt-btcx-aydz=\det(A)\det(B)$, which concludes the proof.

Answer 2

$\det(-)$ is the unique alternating multilinear function from $\Bbb R^2\times \Bbb R^2\to \Bbb R$ for which $\det(e_1,e_2)=1$. In particular, if $f: \Bbb R^2\times \Bbb R^2\to \Bbb R$ is multilinear alternating and $f(e_1,e_2)=\alpha$, $f=\alpha\det$. Show that for fixed $A$, $f(-)=\det(A-)$ is multilinear alternating and $f(e_1,e_2)=\det(A)$.