Question
Question: If A<sup>k</sup> = 0, for some value of k, (I – A)<sup>p</sup> = I + A + A<sup>2</sup> + …. A<sup>k–...
If Ak = 0, for some value of k, (I – A)p = I + A + A2 + …. Ak–1, thus p is (A is nilpotent with index k).
A
–1
B
–2
C
–3
D
None of these
Answer
–1
Explanation
Solution
Let B = I + A + A2 + … + Ak–1
Post multiply both sides by (I – A), so that
B(I – A) = (I + A + A2 + … + Ak–1) (I – A)
= I – A + A – A2 + A2 – A3 + … –Ak–1 + Ak–1 – Ak
= I – Ak = I, since Ak = 0
Ž B = (I – A)–1
Hence (I – A)–1 = I + A + A2 + … + Ak–1.
Thus, p = –1.
Hence (1) is correct answer.