Calculating the inverse modularity of the determinant for Hill cipher I'm trying to decrypt a message encrypted with Hill Cipher, but I don't understand how to find the determinant so it solves the equation $det * 1/det = 1 mod 26$.

The determinant for my key matrix is $62$.
Maybe the impossibility of solving the equation system uniquely was meant to strengthen the cipher. :D If everything was done correctly, there'll be multiple solutions. When you have some idea of how the plaintext may look like, it should be easy to determine it uniquely.

During the computation you have to divide something by 62 modulo 26, which is (as already stated) impossible. However, the nominator will always be even, and for any $a$, finding a solution to

$$62 * x \equiv 2a \pmod{26}$$

is as easy as finding the unique (mod 13) solution to

$$31 * x \equiv a \pmod{13}$$

Just note that for each $x$ also $x' = x+13$ solves the original equation.