Why does $|{ad-bc}| \not =0$ imply that the transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$, given by $T(x,y) = (g(x,y),h(x,y))$ where $g(x,y)=ax+by$ and $h(x,y)=cx+dy$, is a bijection
I can see that is if ad-bc=o then the T isn't invertible(which is of course equivalent to being bijective) but I would like to know : 1) how this condition implies that T is surjective and how it implies that T is injective 2) and why is there a need for the modulus(absolute value) around ad-bc?
Your linear transformation $T$ is encoded as matrix
$\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$
And by linear algebra such trafo is bijective iff it's determinant $det(A)$ isn't zero. Since $det(T)= ad-bc$ this is equivalent to $|{ad-bc}| \not =0$