<s>[INST] 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!
 [/INST] Notice that
$$\begin{align}\begin{bmatrix}a & b\ |\ 1 & 0 \\c & d \ |\  0 & 1\end{bmatrix} &\Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & \frac{ad - bc}{a} \ &|\  -\frac{c}{a} & 1\end{bmatrix} \Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & 1 \ &|\  -\frac{c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\\&\Leftrightarrow \begin{bmatrix}a & 0\ &|\ \frac{ad}{ad-bc} & \frac{-ba}{ad-bc} \\0 & 1 \ &|\  \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix} \Leftrightarrow \begin{bmatrix}1 & 0\ &|\ \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\0 & 1 \ &|\  \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\end{align}$$
Then
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
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