Let P=begin{bmatrix} 1 & 0 & 0 4 & 1 & 0 16 | vedclass

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(B) Given $P = \begin{bmatrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 16 & 4 & 1 \end{bmatrix}$. Let $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \ 4 & 0 & 0

Vedclass · Accepted Answer

$103$

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(B) Given $P = \begin{bmatrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 16 & 4 & 1 \end{bmatrix}$. Let $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \ 4 & 0 & 0 \ 16 & 4 & 0 \end{bmatrix}$.Note that $A^2 = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 16 & 0 & 0 \end{bmatrix}$ and $A^3 = O$ (zero matrix).Using the Binomial Theorem,$P^n = (I+A)^n = I + nA + \frac{n(n-1)}{2}A^2$.For $n=50$,$P^{50} = I + 50A + \frac{50 	imes 49}{2}A^2 = I + 50A + 1225A^2$.$P^{50} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + 50 \begin{bmatrix} 0 & 0 & 0 \ 4 & 0 & 0 \ 16 & 4 & 0 \end{bmatrix} + 1225 \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 16 & 0 & 0 \end{bmatrix}$.$P^{50} = \begin{bmatrix} 1 & 0 & 0 \ 200 & 1 & 0 \ 800 + 19600 & 200 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 200 & 1 & 0 \ 20400 & 200 & 1 \end{bmatrix}$.Since $P^{50}-Q=I$,we have $Q = P^{50}-I = \begin{bmatrix} 0 & 0 & 0 \ 200 & 0 & 0 \ 20400 & 200 & 0 \end{bmatrix}$.Thus,$q_{31} = 20400$,$q_{32} = 200$,and $q_{21} = 200$.Therefore,$\frac{q_{31}+q_{32}}{q_{21}} = \frac{20400+200}{200} = \frac{20600}{200} = 103$.

Let $P=\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix}$ and $I$ be the identity matrix of order $3$. If $Q=[q_{ij}]$ is a matrix such that $P^{50}-Q=I$,then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

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