If A=[\table 1 , 2 ; 3 , 4], then I+A+A^{2}+. . . \infty equals to

Correct Answer is : [12−13−120]

[\table \frac{1}{2} , -\frac{1}{3} ; -\frac{1}{2} , 0]

[\table -\frac{1}{4} , \frac{1}{3} ; \frac{1}{2} , 0]

NIMCET 2009 Question Paper with solutions

Question : 18 of 120

Marks: +1, -0

A=[\table 1 , 2 ; 3 , 4]

, then

I+A+A^{2}+. . . ∞

equals to

Solution:

Let

λ

be an eigen value of given matrix

A

.
Now,

\;\; 1+λ+λ^{2}+. . . α=\;{1}/{1-λ}

⇒ \;\; (1-λ)(1+λ+λ^{2}+. . . ∞)=1

by Cayley-Hamilton's theorem, "Every square matrix satisfy its characteristic equation".

(I-A)(I+A+A^{2}+. . . ∞)=I

⇒ I +A+A^{2}+. . .+∞=(I-A)^{-1}

=[\table 0 , -2 ; -3 , -3]^{-1}=-\;{1}/{6}[\table -3 , 2 ; 3 , 0]

=[\table \;{1}/{2} , -\;{1}/{3} ; -\;{1}/{2} , 0]

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