I've been trying to find the inverse of a matrix through this sort of roundabout manner but to no avail. Sure, I could just use the standard way, but I figured working through this particular process would be more intuitive. Anyway, my work goes as follows:
$$AA^{-1} = A^{-1}A=I $$
$$ \text{Let A be} = \pmatrix{2&3\\1&2}$$
$$\pmatrix{2&3\\1&2}\pmatrix{a&b\\c&d} = \pmatrix{a&b\\c&d}\pmatrix{2&3\\1&2}=\pmatrix{1&0\\0&1}$$
Expanding gives:
$$\pmatrix{2a+3c&2b+3d\\a+2c&b+2d} = \pmatrix{2a+b&3a+2b\\2c+d&3c+2d}=\pmatrix{1&0\\0&1}$$
$$\therefore 2a+3c=2a+b=0$$ $$\therefore 2b+3d=3a+2b=0$$
Solving gives:
$$a = 0, b=1, c=\frac{1}{3}, d=0$$
Hence the derived inverse is:
$$\pmatrix{0&1\\\frac{1}{3}&0}$$
However, this is not the inverse of the A matrix above. Where is the flaw in this method?
it seems there's a typo in your system, as : $$2a+3c=2a+b=1$$ and not $0$.
Solving this (and checked it with WolframAlpha) gave me $a=2, b=-3, c=-1, d=2$, which seems to be correct.
The system works fine, so good job on the method!