Determinants Matrices Question 115
Question: If $ A^{k}=0 $ (A is nilpotent with index k), $ {{(I-A)}^{p}}=I+A+A^{2}+….+{A^{k-1}}, $ thus p is,
Options:
A) -1
B) -2
C) ½
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Let $ B=I+A+A^{2}+….+{A^{k-1}} $
Now multiply both sides by (I - A), we get $ B(I-A)=(I+A+A^{2}+….+{A^{k-1}}(I-A) $
$ =I-A+A-A^{2}+A^{2}-A^{3}+….-{A^{k-1}}+{A^{k-1}}-A^{k} $
$ =I-A^{k}=I, $ since $ A^{k}=0\Rightarrow B={{(I-A)}^{-1}} $ Hence $ {{(I-A)}^{-1}}=I+A+A^{2}+…+{A^{k-1}} $ Thus $ p=-1 $
BETA
