Question-47
If:
\[ \begin{equation*} A=\begin{bmatrix} 1 & 0 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 1 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 1 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3 \end{bmatrix} \end{equation*} \]
Then consider the following matrices:
\[ \begin{equation*} \begin{aligned} P & =\begin{bmatrix} 1 & 0 & 0 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300}\\ 0 & 1 & 0 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300}\\ 0 & 0 & 1 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300}\\ 0 & 0 & 0 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300}\\ 0 & 0 & 0 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300}\\ 0 & 0 & 0 & ( 1/3)^{300} & ( 1/3)^{300} & ( 1/3)^{300} \end{bmatrix} \end{aligned} ,\ Q=\begin{bmatrix} 1 & 0 & 0 & 100 & 100 & 100\\ 0 & 1 & 0 & 100 & 100 & 100\\ 0 & 0 & 1 & 100 & 100 & 100\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3\\ 0 & 0 & 0 & 1/3 & 1/3 & 1/3 \end{bmatrix} \end{equation*} \]
What is \(\displaystyle A^{300}\) equal to?
Look for an idempotent matrix.
This can be partitioned as follows:
\[ \begin{equation*} A=\begin{bmatrix} I & B\\ 0 & B \end{bmatrix} \end{equation*} \]
where, \(\displaystyle B=\frac{1}{3}\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}\). Notice that \(\displaystyle B^{2} =B\).
First, let us compute \(\displaystyle A^{2}\):
\[ \begin{equation*} A^{2} =\begin{bmatrix} I & B\\ 0 & B \end{bmatrix}\begin{bmatrix} I & B\\ 0 & B \end{bmatrix} =\begin{bmatrix} I & 2B\\ 0 & B \end{bmatrix} \end{equation*} \]
Assuming that \(\displaystyle A^{n-1} =\begin{bmatrix} I & ( n-1) B\\ 0 & B \end{bmatrix}\), we have: \[ \begin{equation*} A^{n} =\begin{bmatrix} I & ( n-1) B\\ 0 & B \end{bmatrix}\begin{bmatrix} I & B\\ 0 & B \end{bmatrix} =\begin{bmatrix} I & nB\\ 0 & B \end{bmatrix} \end{equation*} \]
Hence we have shown that for all \(n \geqslant 2\):
\[ \begin{equation*} A^{n} =\begin{bmatrix} I & nB\\ 0 & B \end{bmatrix} \end{equation*} \]