What are the values of b such that the matrix [(1,1)(b,1)] is invertible mod 26.
I figured that the matrix is only invertible if its determinant and the n value 26 's gcd is 1, meaning they are relatively prime.
So, det(matrix) = (1-b), and then for all values gcd((1-b),n) = 1 would conclude that that's the values of b that are valid.
But would this yield an unlimited amount of values of b?
You must have $b-1\ne 0 \pmod{26}$ and the restrictions $b-1\ne 2n$ and $b-1\ne 13m$ because $b-1\ne 0\pmod{2}$ and $b-1\ne 0\pmod{13}$. In other words $b$ even and $b-1$ not multiple of $13$.