I tried to prove the inverse of the matrix but I got a wrong formula. Here's the proof:
Let $$aX + bY = m \quad , \quad cX + dY = n\\ \begin{align}X & = \frac{m - by}{a}\\ & = \frac{n - dy}{c}\end{align}\\ \therefore cm - cbY = an - adY\\ cm - an = Y(cb - ad)\\\therefore Y = \frac{cm - an}{cb - ad}$$ Similarly, $$\therefore X = \frac{dm - bn}{da - cb} $$
If $B$ is inverse of matrix $A$, then $AB = BA = I$
Let $A = \begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix}$ and $B = \begin{pmatrix}b_1 & b_2 \\ b_3 & b_4\end{pmatrix}$
Multiplication forms a series of equations, like: $a_1b_1 + a_2b_3 = 1$ and $ a_1b_2 + a_2b_4 = 1$
But there is clearly something wrong here because according to 1, then $b_1 = b_2$ and $b_3 = b_4$ which is not always true. What's wrong with this proof ?
Your second equation is wrong, hence the apparent contradiction.
You have the first row of $A$ multiplied by the second column of $B$. The result of that multiplication therefore gives the first row, second column entry of $AB=I$. The rest of the correction is trivial.