Question

Let 'a' be a matrix with eigen value 1,-1,0 then the determinant of i+a^100

Answer 1

let a b c be Eugen values of new determinant
a=i+(1)^100=i+1
b=i+(-1)^100=i+1
c=i
Now determinant is given by abc
=(1+i)(1+i)i
(i-1)(1+i)
=-1-1=-2