Question

Let $P_{n}$ be a square matrix of order $3$ such that $P_{n}=\left[a_{i j}\right]$ , where $a_{i j}=\frac{3 i + j}{4^{2 n}}$ for $1\leq i\leq 3,1\leq j\leq 3.$ Then the value of $\underset{n \rightarrow \in fty}{l i m}T_{r}\left(4 P_{1} + 4^{2} P_{2} . . . . . 4^{n} P_{n}\right)$ 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)

$\because P_{n}=\left[\frac{3 i+j}{4^{2 n}}\right]=\frac{1}{4^{2 n}}[3 i+j]$
$4^{n} P_{n}=\frac{1}{4^{n}}[3 i+j]$
$T_{r}\left(4 P_{1}+4^{2} P_{2} \ldots 4^{n} P_{n}\right)=T_{r}\left(4 P_{1}\right)+T_{r}\left(4^{2} P_{2}\right) \ldots+T_{r}\left(4^{n} P_{n}\right)$
$=\frac{24}{4}+\frac{24}{4^{2}} \ldots \ldots \ldots \frac{24}{4^{n}}$
$\lim _{n \rightarrow \infty}\left(T_{r}\left(4 P_{1}+4^{2} P_{2} \ldots+4^{n} P_{n}\right)\right)$
$\lim _{n \rightarrow \infty}\left(\frac{24}{4}+\frac{24}{4^{2}} \ldots \ldots . \frac{24}{4^{n}}\right)$
$=\frac{6}{1-\frac{1}{4}}=8$