Q. Let

A = ⎡ 3 −4 ⎤
  ⎣ 1 −1 ⎦

and B be two matrices such that A¹⁰⁰ = 100B + I.

Then the sum of all the elements of B¹⁰⁰ is ______.

Correct Answer: 0

Explanation

First find the characteristic equation of matrix A.

|A − λI| = | 3−λ −4 |
     | 1 −1−λ |

= (3−λ)(−1−λ) + 4

= λ² − 2λ + 1

= (λ − 1)²

So, A has a repeated eigenvalue λ = 1.

Hence, A can be written as:

A = I + N

where N is a nilpotent matrix.

Then,

A¹⁰⁰ = (I + N)¹⁰⁰ = I + 100N

Comparing with:

A¹⁰⁰ = 100B + I

we get:

B = N

Since N is nilpotent of order 2,

N² = 0

Therefore,

B¹⁰⁰ = 0

Hence, the sum of all elements of B¹⁰⁰ is:

0

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