Matrices and Probability question Question: 

Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that  $A^2B=A$ . Given $S=\{-1,0,1\}$ and 
    $$ A=   \begin{bmatrix} a & b  \\   c & d \\    \end{bmatrix} $$
  in which $a,b,c,d\in S $ . What is the probability that $det(A+B)=detA + detB $

I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $\frac{8}{81}$? But i'm not sure if its the correct answer.
 A: Yes, you are correct!
Since $A$ is non-singular:
$$A^2B=A \iff AB=I \iff \\
B=A^{-1}=\frac{1}{ad-bc}
\begin{pmatrix} 
d & -b\\
-c & a
\end{pmatrix};\\
A+B=\begin{pmatrix}a+\frac{d}{ad-bc}&b-\frac{b}{ad-bc}\\ c-\frac{c}{ad-bc}&d+\frac{a}{ad-bc}\end{pmatrix}.$$
So:
$$\det(A)+\det(B)=ad-bc+\frac{1}{(ad-bc)^2}(ad-bc)=\frac{(ad-bc)^2+1}{ad-bc};\\
\det(A+B)=ad+\frac{a^2+d^2}{ad-bc}+\frac{ad}{(ad-bc)^2}-bc+\frac{2bc}{ad-bc}-\frac{bc}{(ad-bc)^2};\\
\det(A)+\det(B)=\det(A+B) \Rightarrow \\
(ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 \Rightarrow \\
a^2+d^2=-2bc.$$
Note that $ad\ne bc$ and the RHS must be positive. Then the favorable outcomes are:
$$\begin{array}{r|r|r|r}
N&a&b&c&d\\
\hline
1&-1&-1&1&-1\\
2&1&-1&1&1\\
3&-1&1&-1&-1\\
4&1&1&-1&1
\end{array}$$
There are total $3^4$ possible ways for $a,b,c,d\in \{-1,0,1\}$. 
Hence the required probability is:
$$P=\frac{n(\text{favorable})}{n(\text{total})}=\frac4{81}.$$
A: Use $$\det(A)\det(A+B)= \det (A)(\det(A)+\det(B))$$
so $$ \det (A(A+B)) = \det (A^2)+\det (AB)$$
so $$ \det (A^2+I) = \det (A^2)+1$$
since $$ A^2=   \begin{bmatrix} a^2+bc & b(a+d)  \\   c(a+d) & d^2+bc \\    \end{bmatrix} = \begin{bmatrix} x & y  \\   z & t \\    \end{bmatrix}$$
we get $$(x+1)(t+1)-yz = xt-yz+1\implies x+t=0$$
so $$a^2+d^2+2bc=0$$
