The correct option is A Eigen value 3 has a multiplicity of 2 and only one independent eigen vector exists
[A] = [5−141]
For eigen value
|A−λI| = ∣∣∣5−λ−141−λ∣∣∣ = 0
⇒ (5−λ)(1−λ)+4 = 0
5−5λ−λ+λ2+4 = 0
λ2−6λ+9 = 0
(λ−3)2 = 0
λ = 3, 3
So it has multiplicity 'two'
For eigen vector
[A−λI][xy] = 0
[5−3−141−3][xy]=[00]
2x - y = 0 ⇒ y = 2x.
Let x = k, then y = 2k
So X = [xy] = [k2k] ∼[12]
So, only one independent eigen vector exists for λ = 3.