Question

Let $P_n$ be a square matrix of order $3$ such that $P_n=\left[a_{ij}\right]$ , where $a_{ij}=\frac{3i+j}{4^{2n}}$ for $1\le i\le 3,1\le j\le 3.$ Then the value of $\underset{n\to \in fty}{\lim }{T}_r\left(4P_1+4^2{P}_2.....4^n{P}_n\right)$ is

(where $T_r\left(A\right)$ denotes trace of matrix $A$ i.e sum of principal diagonal elements of $A$ )

• A. 7
• B. 8
• C. $\frac{25}{3}$
• D. $9$

Answer 1

Answer: (B)

$\because P_n=\left[\frac{3i+j}{4^{2n}}\right]=\frac{1}{4^{2n}}[3i+j]$
$4^n{P}_n=\frac{1}{4^n}[3i+j]$
$T_r\left(4P_1+4^2{P}_2\dots 4^n{P}_n\right)=T_r\left(4P_1\right)+T_r\left(4^2{P}_2\right)\dots +T_r\left(4^n{P}_n\right)$
$=\frac{24}{4}+\frac{24}{4^2}\dots \dots \dots \frac{24}{4^n}$
${\lim }_{n\to \infty }\left(T_r\left(4P_1+4^2{P}_2\dots +4^n{P}_n\right)\right)$
${\lim }_{n\to \infty }\left(\frac{24}{4}+\frac{24}{4^2}\dots \dots .\frac{24}{4^n}\right)$
$=\frac{6}{1-\frac{1}{4}}=8$