Diagonalisation and powers Problem set B Answers

1. (a) \(P = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix},\; D = \begin{pmatrix}7 & 0 \\ 0 & 4\end{pmatrix}\)
2. (b) \(P = \begin{pmatrix}1 & 3 \\ 1 & -2\end{pmatrix},\; D = \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}\)

\(M^n = \dfrac{1}{3}\begin{pmatrix}2(7^n) + 4^n & 7^n - 4^n \\ 2(7^n) - 2(4^n) & 7^n + 2(4^n)\end{pmatrix}\); \(M^4 = \begin{pmatrix}1686 & 715 \\ 1430 & 971\end{pmatrix}\)

\(P = \begin{pmatrix}3 & 1 \\ 4 & -1\end{pmatrix},\; D = \begin{pmatrix}6 & 0 \\ 0 & -1\end{pmatrix},\; M^5 = \begin{pmatrix}3332 & 3333 \\ 4444 & 4443\end{pmatrix}\)

1. (a) \(M = \begin{pmatrix}-4 & 3 \\ -18 & 11\end{pmatrix}\)
2. (b) \(M^{3} = \begin{pmatrix}-226 & 117 \\ -702 & 359\end{pmatrix}\)

(a) \(T^3\begin{pmatrix}10000 \\ 10000\end{pmatrix} = \begin{pmatrix}8250 \\ 11750\end{pmatrix}\). (b) \(T^n \to \begin{pmatrix}0.4 & 0.4 \\ 0.6 & 0.6\end{pmatrix}\), so 8000 on the first network and 12 000 on the second.

1. (a) \(M = \begin{pmatrix}0 & 1.6 \\ 0.3 & 0.8\end{pmatrix}\)
2. (b) \(\lambda = 1.2 \text{ or } -0.4\)
3. (c) 20% a year; 57.1% are young.

\(M^3 = (PDP^{-1})(PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)D(P^{-1}P)DP^{-1} = PD^3P^{-1}\). Each \(P^{-1}P\) in the middle is \(I\), and this happens \(n - 1\) times for \(M^n\).

\(0.7^n \to 0\), so \(D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\). With 1.3, \(1.3^n\) grows without limit, so \(M^n\) does not settle.