If $ a + d\equiv 1$ then by the first and sixth: $ ad - bc \equiv ad - (a - a^2) \equiv a(d + a - 1) \equiv 0$ contradiction! So $ a + d$ is not congruent to 1, so $ b\equiv c\equiv 0$ $ a(a - 1) \equiv 0$ $ d(d - 1) \equiv 0$ $ ad \equiv 1$ So by the third, neither a nor b is congruent to 0, so they're both congruent to 1. Hence, $ a\equiv\b\equiv 1$ And we're done.

I think I got another solution: Consider $ 2 \times 2$ matrix$ A=$\[ \left( \begin{array}{ccc}a & b \\c & d \\\end{array} \right)\] From first four equation, it gives that $ A^2=A$, From last equation, it dives that $ det(A)=1(\neq0)$. It means $ A^{-1}=$\[ \left( \begin{array}{ccc}d & -b \\-c & a \\\end{array} \right)\] Then, $ A^2A^{-1}=AA^{-1} \Rightarrow A=I$ So, $ a=d=1$ and $ b=c=0$