Diagonalisation and powers Problem set A Answers

1. (a) \(P = \begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix}\)
2. (b) \(P = \begin{pmatrix}1 & 1 \\ 2 & -1\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & -1\end{pmatrix}\)

\(M^n = \dfrac{1}{3}\begin{pmatrix}5^n + 2(2^n) & 2(5^n) - 2(2^n) \\ 5^n - 2^n & 2(5^n) + 2^n\end{pmatrix}\); \(M^4 = \begin{pmatrix}219 & 406 \\ 203 & 422\end{pmatrix}\)

\(P = \begin{pmatrix}1 & 3 \\ 1 & -4\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & -2\end{pmatrix},\; M^5 = \begin{pmatrix}1772 & 1353 \\ 1804 & 1321\end{pmatrix}\)

1. (a) \(M = \begin{pmatrix}-2 & 3 \\ -6 & 7\end{pmatrix}\)
2. (b) \(M^{3} = \begin{pmatrix}-62 & 63 \\ -126 & 127\end{pmatrix}\)

(a) \(T^3\begin{pmatrix}5000 \\ 5000\end{pmatrix} = \begin{pmatrix}5875 \\ 4125\end{pmatrix}\). (b) \(T^n \to \begin{pmatrix}0.6 & 0.6 \\ 0.4 & 0.4\end{pmatrix}\), so 6000 in the town and 4000 in the countryside.

1. (a) \(M = \begin{pmatrix}0 & 2.2 \\ 0.2 & 0.7\end{pmatrix}\)
2. (b) \(\lambda = 1.1 \text{ or } -0.4\)
3. (c) 10% a year; 66.7% are young.

\(M^n = (PDP^{-1})(PDP^{-1})\cdots(PDP^{-1})\). Each \(P^{-1}P\) in the middle is \(I\), leaving \(PD^nP^{-1}\).

\(0.4^n \to 0\), so \(D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\). With 1.4, \(1.4^n\) grows without limit, so \(M^n\) does not settle.