Question

Matrix question.

Answer 1

If \(\Huge [A_k]_{n x n}\) is a square matrix and \(\Huge a_{ij}=0\) if \(\Huge i \neq j\) and \(\Huge a_{ij}=\frac{1}{k+i}\) if \(\Huge i=j\) & \(\Huge [B_k]_{n x n}\) is inverse of \(\Huge [A_k]_{n x n}\) then answer the following question.
Find
\[\Huge \lim_{m \rightarrow \infty} \frac{\sum_{n=1}^{m} trace[B_k]_{n \times n}}{m^3}\]

Answer 2

that is n cross n :) not nxn :|

Answer 3

@dan815 @ganeshie8 @genius12 @oldrin.bataku @terenzreignz @mukushla

Answer 4

\[\Huge \left[\begin{matrix}\frac{1}{k+1} & 0 & 0 & 0...\\ 0 & \frac{1}{k+2}& 0 &0.... \\0 & 0 & \frac{1}{k+3}...\end{matrix}\right]\]
I guess something like this will be created

Answer 5

\[\Huge [B]=\left[\begin{matrix}k+1 & 0 & 0 & 0...\\ 0 & k+2 &0 &0...\end{matrix}\right]\]

Answer 6

\[\LARGE Trace=(k+1)+(k+2)+(k+3)...\]

Answer 7

Need some help with this part..my summation and stuff part isn't strong..

Answer 8

@hartnn

Answer 9

So you first need to calculate \[\mbox{tr}(B_k)=(k+1)+(k+2)+\cdots +(k+n)\]

Answer 10

Thats going to be:\[(k+1)+(k+2)+\cdots +(k+n)\]\[=nk+(1+2+3+\cdots +n)\]\[=nk+\frac{1}{2}n(n+1).\]

Answer 11

yup

Answer 12

Now you can calculate: \[\sum_{n=1}^m\mbox{tr}(B_k)_{n\times n}\]\[=\sum_{n=1}^m nk+\frac{1}{2}n(n+1)=\sum_{n=1}^m\frac{1}{2}n^2+(k+\frac{1}{2})n\]\[=\frac{1}{2}\sum_{n=1}^mn^2+\left(k+\frac{1}{2}\right)\sum_{n=1}^mn\]\[=\frac{1}{2}\cdot \frac{m(m+1)(2m+1)}{6}+\left(k+\frac{1}{2}\right)\cdot\frac{m(m+1)}{2}.\]

Answer 13

Notice that this will be a cubic function in terms of \(m\), so in calculating:\[\lim_{m\rightarrow \infty}\frac{\sum_{n=1}^m\mbox{tr}(B_k)_{n\times n}}{m^3}\]we only care about what the coefficient of the \(m^3\) is on the top part of the fraction.

Answer 14

1/6 it would be ?

Answer 15

The coefficient of the cubic polynomial in \(m\) will be \(\frac{2}{12}=\frac{1}{6}.\) So your limit will be \(\frac{1}{6}.\)

Answer 16

thank you sorry to intrude.. .it was a beautiful solution :)

Answer 17

Thank you :)

Answer 18

amazing,thanks!