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IB Mathematics: Applications and Interpretation HL · AHL 5: Matrices · AHL 5.7

Diagonalisation and powers Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Write each matrix in the form \(PDP^{-1}\).

    1. (a) \(\begin{pmatrix}6 & 1 \\ 2 & 5\end{pmatrix}\)

      • \(\det(M - \lambda I) = (6 - \lambda)(5 - \lambda) - (1)(2) = 0\)
      • \(\lambda^2 - 11\lambda + 28 = 0\), so \((\lambda - 7)(\lambda - 4) = 0\): \(\lambda = 7\) or \(\lambda = 4\).
      • For \(\lambda = 7\): \((M - 7I)v = 0\) gives \(-x + y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
      • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(2x + y = 0\), so \(v = \begin{pmatrix}1 \\ -2\end{pmatrix}\).
      • \(P = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix},\; D = \begin{pmatrix}7 & 0 \\ 0 & 4\end{pmatrix}\), so \(M = PDP^{-1}\).

    2. (b) \(\begin{pmatrix}1 & 3 \\ 2 & 2\end{pmatrix}\)

      • \(\det(M - \lambda I) = (1 - \lambda)(2 - \lambda) - (3)(2) = 0\)
      • \(\lambda^2 - 3\lambda - 4 = 0\), so \((\lambda - 4)(\lambda + 1) = 0\): \(\lambda = 4\) or \(\lambda = -1\).
      • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(-3x + 3y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
      • For \(\lambda = -1\): \((M + I)v = 0\) gives \(2x + 3y = 0\), so \(v = \begin{pmatrix}3 \\ -2\end{pmatrix}\).
      • \(P = \begin{pmatrix}1 & 3 \\ 1 & -2\end{pmatrix},\; D = \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}\), so \(M = PDP^{-1}\).

  2. For \(M = \begin{pmatrix}6 & 1 \\ 2 & 5\end{pmatrix}\), find \(M^n\) in terms of \(n\), and hence find \(M^4\).

    • Eigenvalues 7 and 4 with eigenvectors \(\begin{pmatrix}1 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -2\end{pmatrix}\), so \(P = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix},\; D = \begin{pmatrix}7 & 0 \\ 0 & 4\end{pmatrix}\).
    • \(P^{-1} = \frac{1}{-3}\begin{pmatrix}-2 & -1 \\ -1 & 1\end{pmatrix} = \frac{1}{3}\begin{pmatrix}2 & 1 \\ 1 & -1\end{pmatrix}\).
    • \(M^n = PD^nP^{-1} = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix}\begin{pmatrix}7^n & 0 \\ 0 & 4^n\end{pmatrix}\cdot\frac{1}{3}\begin{pmatrix}2 & 1 \\ 1 & -1\end{pmatrix}\)
    • \(= \frac{1}{3}\begin{pmatrix}2(7^n) + 4^n & 7^n - 4^n \\ 2(7^n) - 2(4^n) & 7^n + 2(4^n)\end{pmatrix}\)
    • For \(n = 4\), \(7^4 = 2401\) and \(4^4 = 256\): \(M^4 = \frac{1}{3}\begin{pmatrix}5058 & 2145 \\ 4290 & 2913\end{pmatrix} = \begin{pmatrix}1686 & 715 \\ 1430 & 971\end{pmatrix}\).

  3. \(M = \begin{pmatrix}2 & 3 \\ 4 & 3\end{pmatrix}\) has eigenvalues 6 and −1. Diagonalise \(M\) and use this to find \(M^5\).

    • For \(\lambda = 6\): \(-4x + 3y = 0\), so \(v = \begin{pmatrix}3 \\ 4\end{pmatrix}\). For \(\lambda = -1\): \(3x + 3y = 0\), so \(v = \begin{pmatrix}1 \\ -1\end{pmatrix}\).
    • \(P = \begin{pmatrix}3 & 1 \\ 4 & -1\end{pmatrix},\; D = \begin{pmatrix}6 & 0 \\ 0 & -1\end{pmatrix}\).
    • \(M^5 = PD^5P^{-1} = \begin{pmatrix}3 & 1 \\ 4 & -1\end{pmatrix}\begin{pmatrix}7776 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{7} & \frac{1}{7} \\ \frac{4}{7} & -\frac{3}{7}\end{pmatrix}\)
    • From the calculator: \(M^5 = \begin{pmatrix}3332 & 3333 \\ 4444 & 4443\end{pmatrix}\).

  4. A matrix \(M\) has eigenvalue \(5\) with eigenvector \(\begin{pmatrix}1 \\ 3\end{pmatrix}\) and eigenvalue \(2\) with eigenvector \(\begin{pmatrix}1 \\ 2\end{pmatrix}\).

    1. (a) Find \(M\).

      • \(P = \begin{pmatrix}1 & 1 \\ 3 & 2\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix},\; P^{-1} = \begin{pmatrix}-2 & 1 \\ 3 & -1\end{pmatrix}\)
      • \(M = PDP^{-1} = \begin{pmatrix}5 & 2 \\ 15 & 4\end{pmatrix}\begin{pmatrix}-2 & 1 \\ 3 & -1\end{pmatrix} = \begin{pmatrix}-4 & 3 \\ -18 & 11\end{pmatrix}\)

    2. (b) Find \(M^{3}\).

      • \(M^{3} = PD^{3}P^{-1} = \begin{pmatrix}1 & 1 \\ 3 & 2\end{pmatrix}\begin{pmatrix}125 & 0 \\ 0 & 8\end{pmatrix}\begin{pmatrix}-2 & 1 \\ 3 & -1\end{pmatrix}\)
      • \(= \begin{pmatrix}-226 & 117 \\ -702 & 359\end{pmatrix}\)

In context

  1. Each year 30% of the customers of one phone network switch to its rival, and 20% of the rival's customers switch to it. This is modelled by \(T = \begin{pmatrix}0.7 & 0.2 \\ 0.3 & 0.8\end{pmatrix}\). At the start there are 10 000 customers on each network. (a) Find the numbers after 3 years. (b) Find the matrix that \(T^n\) approaches, and the numbers in the long term.

    • (a) From the calculator: \(T^3 = \begin{pmatrix}0.475 & 0.35 \\ 0.525 & 0.65\end{pmatrix}\).
    • \(T^3\begin{pmatrix}10000 \\ 10000\end{pmatrix} = \begin{pmatrix}8250 \\ 11750\end{pmatrix}\).
    • (b) The eigenvalues are 1 and 0.5, with eigenvectors \(\begin{pmatrix}2 \\ 3\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -1\end{pmatrix}\), so \(P = \begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix},\; D = \begin{pmatrix}1 & 0 \\ 0 & 0.5\end{pmatrix}\).
    • \(D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\), so \(T^n \to P\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}P^{-1} = \begin{pmatrix}0.4 & 0.4 \\ 0.6 & 0.6\end{pmatrix}\). Then \(\begin{pmatrix}0.4 & 0.4 \\ 0.6 & 0.6\end{pmatrix}\begin{pmatrix}10000 \\ 10000\end{pmatrix} = \begin{pmatrix}8000 \\ 12000\end{pmatrix}\).

  2. Each year every adult seal produces 1.6 pups, 30% of the pups survive to become adults, and 80% of the adults survive. Let \(y_n\) and \(a_n\) be the numbers of young and adult seals in year \(n\).

    1. (a) Write down the matrix \(M\) with \(\begin{pmatrix}y_{n+1} \\ a_{n+1}\end{pmatrix} = M\begin{pmatrix}y_n \\ a_n\end{pmatrix}\).

      • New young come from adults: \(y_{n+1} = 1.6\,a_n\). New adults: \(a_{n+1} = 0.3\,y_n + 0.8\,a_n\)
      • \(M = \begin{pmatrix}0 & 1.6 \\ 0.3 & 0.8\end{pmatrix}\)

    2. (b) Find the eigenvalues of \(M\).

      • \(\lambda^2 - 0.8\lambda - 0.48 = 0\)
      • \(\lambda = 1.2 \text{ or } -0.4\)

    3. (c) Find the long-term percentage increase in the population each year, and the long-term percentage of the seals that are young.

      • The larger eigenvalue \(1.2\) gives the long-term multiplier: an increase of 20%.
      • Eigenvector: \(\begin{pmatrix}4 \\ 3\end{pmatrix}\)
      • Share that are young: \(\dfrac{4}{4+3} = 0.571\), which is 57.1%.

Stretch

  1. \(M = PDP^{-1}\). Show that \(M^3 = PD^3P^{-1}\), and explain why the same works for any \(n\).

    • \(M^3 = (PDP^{-1})(PDP^{-1})(PDP^{-1})\).
    • Each \(P^{-1}P\) in the middle is \(I\), so \(M^3 = PD(I)D(I)DP^{-1} = PD^3P^{-1}\).
    • For \(M^n\) there are \(n - 1\) such pairs, all equal to \(I\), leaving \(PD^nP^{-1}\).

  2. The eigenvalues of a matrix \(M\) are 1 and 0.7. Explain why \(M^n\) settles to a fixed matrix as \(n\) grows. What would happen if they were 1 and 1.3?

    • \(D = \begin{pmatrix}1 & 0 \\ 0 & 0.7\end{pmatrix}\) so \(D^n = \begin{pmatrix}1 & 0 \\ 0 & 0.7^n\end{pmatrix}\).
    • \(0.7^n \to 0\) as \(n\) grows, so \(D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\) and \(M^n = PD^nP^{-1}\) settles.
    • With 1.3, \(1.3^n\) grows without limit, so \(M^n\) does not settle.

jmaths.xyz · AI HL · AHL 5.7 Diagonalisation and powers of matrices