JEE Main On 16 April 2018 Question 1
Question: Let $ A= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} $ and $ B=A^{20}. $ Then the sum of the elements of the first column of B is? [JEE Main 16-4-2018]
Options:
A) 211
B) 210
C) 231
D) 251
Show Answer
Answer:
Correct Answer: C
Solution:
Given, $ A= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} . $ Computing higher powers of A,
$ A^{2}=A.A= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{bmatrix} $
$ A^{3}=A^{2}.A= \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \\ \end{bmatrix} $
$ A^{4}=A^{3}.A= \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} $ $ = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 10 & 4 & 1 \\ \end{bmatrix} . $
On observing the pattern, we come to a conclusion that,
$ A^{k}= \begin{bmatrix} 1 & 0 & 0 \\ k & 1 & 0 \\ \frac{k(k+1)}{2} & k & 1 \\ \end{bmatrix} . $
Therefore, sum of first column of $ A^{20} $ is $ (1+20+\frac{20\times 21}{2})=231. $
Option C is correct.
