Answer is 81
Given that det(A-I)=0 and A, I are 3*3 matrix so there are 3 eigen values $\Lambda1,\Lambda2 and \Lambda3$
Trace(A) =13=($\Lambda1+ \Lambda2+\Lambda3$)
Det(A)=32=($\Lambda1*\Lambda2*\Lambda3$)
Now as det(A-$\Lambda$*I)=0
So $\Lambda$1=1
Substituting it in trace and det(A) we get other two eigen values as 8,4
=8^2+4^2+1^2
=81