A) -1
B) -2
C) ½
D) None of these
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 \[p=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
