|=0 ⇒(2−λ)(−λ)−(−1)(1)=−2λ+λ2+1=0 ⇒λ2−2λ+1=0 ⇒(λ−1)2=0 ⇒λ=1 Since A satisfies (A−I)2=0 ∴A=I+N where N=A−I N=[
1
−1
1
−1
] N2=0 Am=(I+N)m=I+mN Am⋅Am=(I+mN)(I+mN)=I+2mN+m2N2 Since N2=0 ⇒Am2=I+2mN Now putting in given condition I+m2N+I+mN=3I−A−6 A−1=[
0
1
−1
2
] A−6=(A−1)6=I+(−6)N ∴ Putting in (i) (m2+m)N=I−(I−6N) (m2+m)N=6N Since N≠0 ⇒m2+m=6 ⇒m2+m−6=0 ⇒(m−2)(m+3)=0 ⇒m=2,−3 ∴ Number of elements in S is 2
