Deriving inverse matrix formula If matrix $A$ is given with dimensions $2 \times2 $ then, 
A is invertible if, and only, if $ad - bc \neq 0$: 
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
How can this be derived?  
I just need hints, I am not good at properties etc.. of matrices, please help!

2*2 Matrices inverse proof
As $A \times A^{-1}=I$
$\begin{bmatrix} x_{11} &x_{12} \\ x_{21}&x_{22} \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$
$ax_{11}+cx_{12} = 1$
$ax_{21}+cx_{22} = 0$
$bx_{11}+dx_{12} = 0$
$bx_{21}+dx_{22} = 1$
$b(ax_{11}+cx_{12}) = abx_{11}+bcx_{12}=b$
$a(bx_{11}+dx_{12}) = abx_{11}+adx_{12}=0$
$(abx_{11}+adx_{12})-(abx_{11}+bcx_{12})=-b$
$x_{12}(ad-bc)=-b$
Row 1 Column 2 : $x_{12} =$ $-b\over{ad-bc}$
$b(ax_{21}+cx_{22}) = abx_{21}+bcx_{22}=0$
$a(bx_{21}+dx_{22}) = abx_{21}+adx_{22}=a$
$(abx_{21}+adx_{22})-(abx_{21}+bcx_{22})=a$
$x_{22}(ad-bc)=a$
Row 2 Column 2 : $x_{22} = $$a \over ad-bc$
$d(ax_{11}+cx_{12}) = adx_{11}+cdx_{12}=d$
$c(bx_{11}+dx_{12}) = bcx_{11}+cdx_{12}=0$
$(adx_{11}+cdx_{12})-(bcx_{11}+cdx_{12})=d$
$x_{11}(ad-bc)=d$
Row 1 Column 1 : $x_{11} = $$d\over ad-bc$
$d(ax_{21}+cx_{22}) = adx_{21}+cdx_{22}=0$
$c(bx_{21}+dx_{22}) = bcx_{21}+cdx_{22}=c$
$(adx_{21}+cdx_{22})-(bcx_{21}+cdx_{22})=-c$
$x_{21}(ad-bc)=-c$
Row 2 Column 1 : $x_{21} = $$-c \over ad-bc$
As $x_{11}=$$d \over ad-bc$, $x_{12}=$$-b \over ad-bc$, $x_{21}$$=-c \over ad-bc $and $x_{22}=$$a \over ad-bc$
$$
A^{-1}=  \begin{bmatrix} \frac{d}{detA} & \frac{-b}{detA}\\ \frac{-c}{detA}& \frac{a}{detA} \end{bmatrix} =
\frac{1}{detA}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}
$$
My first reply was unreadable due to lack of formatting, hope this new revision could help!
