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IB Mathematics: Applications and Interpretation HL · AHL 5: Matrices · AHL 5.2

Matrix multiplication Problem set A 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. Work out each product, or say why it does not exist.

    1. (a) \(\begin{pmatrix}3 & 1 \\ -2 & 4\end{pmatrix}\begin{pmatrix}2 & 0 \\ 5 & -1\end{pmatrix}\)

      • \(\text{Top-left: } 3(2) + 1(5) = 11;\quad \text{top-right: } 3(0) + 1(-1) = -1\)
      • \(\text{Bottom-left: } (-2)(2) + 4(5) = 16;\quad \text{bottom-right: } (-2)(0) + 4(-1) = -4\)
      • \(\begin{pmatrix}3 & 1 \\ -2 & 4\end{pmatrix}\begin{pmatrix}2 & 0 \\ 5 & -1\end{pmatrix} = \begin{pmatrix}11 & -1 \\ 16 & -4\end{pmatrix}\)

    2. (b) \(\begin{pmatrix}1 & -2 & 3\end{pmatrix}\begin{pmatrix}4 \\ 0 \\ -1\end{pmatrix}\)

      • \(1 \times 3 \text{ times } 3 \times 1 \text{ gives a } 1 \times 1 \text{ matrix.}\)
      • \(1(4) + (-2)(0) + 3(-1) = 1\)
      • \(\begin{pmatrix}1\end{pmatrix}\)

    3. (c) \(\begin{pmatrix}4 \\ 0 \\ -1\end{pmatrix}\begin{pmatrix}1 & -2 & 3\end{pmatrix}\)

      • \(3 \times 1 \text{ times } 1 \times 3 \text{ gives a } 3 \times 3 \text{ matrix. Each entry is a row element times a column element.}\)
      • \(\text{Top row: } 4(1),\; 4(-2),\; 4(3) = 4,\; -8,\; 12\)
      • \(\begin{pmatrix}4 & -8 & 12 \\ 0 & 0 & 0 \\ -1 & 2 & -3\end{pmatrix}\)

    4. (d) \(\begin{pmatrix}2 & 0 & 1 \\ -1 & 3 & 2\end{pmatrix}\begin{pmatrix}1 & 4 \\ 0 & -2 \\ 5 & 1\end{pmatrix}\)

      • \(2(1) + 0(0) + 1(5) = 7;\quad 2(4) + 0(-2) + 1(1) = 9\)
      • \((-1)(1) + 3(0) + 2(5) = 9;\quad (-1)(4) + 3(-2) + 2(1) = -8\)
      • \(\begin{pmatrix}7 & 9 \\ 9 & -8\end{pmatrix}\)

    5. (e) \(\begin{pmatrix}3 & 1 \\ -2 & 4\end{pmatrix}\begin{pmatrix}1 & 4 \\ 0 & -2 \\ 5 & 1\end{pmatrix}\)

      • The first matrix is \(2 \times 2\) and the second is \(3 \times 2\). The number of columns of the first (2) must equal the number of rows of the second (3).
      • It does not exist.

    6. (f) \(\begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\)

      • \(1(0) + 2(1) = 2;\quad 1(1) + 2(0) = 1\)
      • \(3(0) + 4(1) = 4;\quad 3(1) + 4(0) = 3\)
      • \(\begin{pmatrix}2 & 1 \\ 4 & 3\end{pmatrix}\text{ (the columns of the first matrix are swapped)}\)

  2. \(A\) has order 3 × 2, \(B\) has order 2 × 4 and \(C\) has order 4 × 3. Give the order of each product, or say that it does not exist.

    1. (a) \(AB\)

      • \(AB: 3 \times 2 \text{ times } 2 \times 4. \text{ The inner numbers match, so the order is } 3 \times 4.\)

    2. (b) \(BA\)

      • \(BA: 2 \times 4 \text{ times } 3 \times 2. \text{ The inner numbers differ (4 and 3), so it does not exist.}\)

    3. (c) \(BC\)

      • \(BC: 2 \times 4 \text{ times } 4 \times 3. \text{ The inner numbers match, so the order is } 2 \times 3.\)

    4. (d) \(CA\)

      • \(CA: 4 \times 3 \text{ times } 3 \times 2. \text{ The inner numbers match, so the order is } 4 \times 2.\)

    5. (e) \(AC\)

      • \(AC: 3 \times 2 \text{ times } 4 \times 3. \text{ The inner numbers differ (2 and 4), so it does not exist.}\)

    6. (f) \(CB\)

      • \(CB: 4 \times 3 \text{ times } 2 \times 4. \text{ The inner numbers differ (3 and 2), so it does not exist.}\)

  3. \(A = \begin{pmatrix}2 & -1 \\ 3 & 0\end{pmatrix}\) and \(B = \begin{pmatrix}1 & 4 \\ -2 & 5\end{pmatrix}\). Find:

    1. (a) \(AB\)

      • \(2(1) + (-1)(-2) = 4\)
      • \(AB = \begin{pmatrix}4 & 3 \\ 3 & 12\end{pmatrix}\)

    2. (b) \(BA\)

      • \(1(2) + 4(3) = 14\)
      • \(BA = \begin{pmatrix}14 & -1 \\ 11 & 2\end{pmatrix}\)
      • \(AB \ne BA\)

    3. (c) \(A^2\)

      • \(2(2) + (-1)(3) = 1;\quad 2(-1) + (-1)(0) = -2\)
      • \(A^2 = \begin{pmatrix}1 & -2 \\ 6 & -3\end{pmatrix}\)

    4. (d) \(A^2 - 3A + 2I\)

      • \(3A = \begin{pmatrix}6 & -3 \\ 9 & 0\end{pmatrix},\quad 2I = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}\)
      • \(A^2 - 3A + 2I = \begin{pmatrix}1 & -2 \\ 6 & -3\end{pmatrix} - \begin{pmatrix}6 & -3 \\ 9 & 0\end{pmatrix} + \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}\)
      • \(\begin{pmatrix}-3 & 1 \\ -3 & -1\end{pmatrix}\)

  4. Find the unknowns.

    1. (a) \(\begin{pmatrix}x & 2 \\ 1 & y\end{pmatrix}\begin{pmatrix}3 \\ -1\end{pmatrix} = \begin{pmatrix}10 \\ 5\end{pmatrix}\)

      • \(\text{Top row: } 3x - 2 = 10, \text{ so } x = 4.\)
      • \(\text{Bottom row: } 3 - y = 5, \text{ so } y = -2.\)
      • \(x = 4,\; y = -2\)

    2. (b) \(\begin{pmatrix}2 & p \\ 1 & 3\end{pmatrix}\begin{pmatrix}1 & 4 \\ q & 2\end{pmatrix} = \begin{pmatrix}-4 & 14 \\ -5 & 10\end{pmatrix}\)

      • \(\text{Top-left: } 2 + pq = -4, \text{ so } pq = -6.\)
      • \(\text{Bottom-left: } 1 + 3q = -5, \text{ so } q = -2.\)
      • \(\text{Then } p = -6 \div (-2) = 3.\)
      • \(\text{Check top-right: } 8 + 2p = 14. \text{ Yes.}\)
      • \(p = 3,\; q = -2\)

In context

  1. A market stall sells apples, pears and plums by the kilogram. The prices in pounds at two stalls, X and Y, are \(P = \begin{pmatrix}2.40 & 2.60 \\ 4.50 & 4.20 \\ 1.80 & 2.00\end{pmatrix}\) (rows: apples, pears, plums; columns: X, Y). (a) Stall X raises all its prices by 4% and stall Y cuts all its prices by 3%. Write this as a matrix product and find the new prices. (b) Instead, apples go up 10%, pears stay the same and plums go down 5%. Write this as a matrix product and find the new prices.

    • \(\text{(a) Each column is one stall, so the multipliers go on the right: } P\begin{pmatrix}1.04 & 0 \\ 0 & 0.97\end{pmatrix}.\)
    • \(= \begin{pmatrix}2.496 & 2.522 \\ 4.68 & 4.074 \\ 1.872 & 1.94\end{pmatrix}\)
    • \(\text{To the penny, in order of rows: } 2.50,\; 2.52;\; 4.68,\; 4.07;\; 1.87,\; 1.94.\)
    • \(\text{(b) Each row is one fruit, so the multipliers go on the left: } \begin{pmatrix}1.1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.95\end{pmatrix}P.\)
    • \(= \begin{pmatrix}2.64 & 2.86 \\ 4.5 & 4.2 \\ 1.71 & 1.9\end{pmatrix}\)

  2. A cinema shows three films. The numbers of adult and child tickets sold are \(\begin{pmatrix}120 & 45 \\ 80 & 110 \\ 65 & 20\end{pmatrix}\). An adult ticket costs £9.50 and a child ticket costs £6. Use a matrix product to find the takings for each film.

    • \(\text{Takings} = \begin{pmatrix}120 & 45 \\ 80 & 110 \\ 65 & 20\end{pmatrix}\begin{pmatrix}9.5 \\ 6\end{pmatrix}\)
    • \(\text{Film 1: } 120(9.5) + 45(6) = 1410\)
    • \(\text{Film 2: } 80(9.5) + 110(6) = 1420;\quad \text{film 3: } 65(9.5) + 20(6) = 737.5\)
    • \(\begin{pmatrix}1410 \\ 1420 \\ 737.5\end{pmatrix} \text{ pounds}\)

  3. In a survey, the proportions of Year 12 and Year 13 students choosing each of three lunch options were \(P = \begin{pmatrix}0.40 & 0.25 \\ 0.35 & 0.45 \\ 0.25 & 0.30\end{pmatrix}\) (columns: Year 12, Year 13). Two schools have \(N = \begin{pmatrix}180 & 200 \\ 160 & 140\end{pmatrix}\) students (rows: Year 12, Year 13; columns: school 1, school 2). Assuming the proportions hold, find the matrix giving the expected number of each lunch in each school, and the total number of the second option needed.

    • \(P \text{ is } 3 \times 2 \text{ and } N \text{ is } 2 \times 2, \text{ so } PN \text{ is } 3 \times 2 \text{: rows are lunch options, columns are schools.}\)
    • \(\text{Option 1, school 1: } 0.40(180) + 0.25(160) = 112\)
    • \(PN = \begin{pmatrix}112 & 115 \\ 135 & 133 \\ 93 & 92\end{pmatrix}\)
    • \(\text{Second option: } 135 + 133 = 268\)

  4. A workshop builds two models of bike, road and trail. This week it builds \(\begin{pmatrix}6 & 4\end{pmatrix}\) (road, trail). The numbers of frames, wheels and gear sets each model needs are \(\begin{pmatrix}1 & 2 & 1 \\ 1 & 2 & 2\end{pmatrix}\) (rows: road, trail). Two shops charge, in euros, \(\begin{pmatrix}120 & 110 \\ 35 & 38 \\ 40 & 36\end{pmatrix}\) (rows: frame, wheel, gear set; columns: shop 1, shop 2). Find the total cost of the parts at each shop, and say which shop is cheaper and by how much.

    • \(\text{Orders: } (1 \times 2)(2 \times 3)(3 \times 2) \text{ gives } 1 \times 2.\)
    • \(OD = \begin{pmatrix}10 & 20 & 14\end{pmatrix}\)
    • \(ODP = \begin{pmatrix}2460 & 2364\end{pmatrix}\)
    • Shop 1 costs €2460.00 and shop 2 costs €2364.00, so shop 2 is cheaper by €96.00.

Stretch

  1. Find the value of \(k\) for which \(\begin{pmatrix}2 & 1 \\ 0 & 3\end{pmatrix}\) and \(\begin{pmatrix}k & 5 \\ 0 & 4\end{pmatrix}\) commute.

    • \(AB = \begin{pmatrix}2 k & 14 \\ 0 & 12\end{pmatrix},\quad BA = \begin{pmatrix}2 k & k + 15 \\ 0 & 12\end{pmatrix}\)
    • \(\text{They commute when } AB = BA. \text{ Only the top-right entries can differ: } 14 = k + 15.\)
    • \(k = -1\)

  2. \(A = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}\). Find \(A^2\) and \(A^3\), and suggest a formula for \(A^n\).

    • \(A^2 = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}\)
    • \(A^3 = A^2 A = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 3 \\ 0 & 1\end{pmatrix}\)
    • \(\text{The top-right entry goes up by 1 each time: } A^n = \begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}.\)

  3. A student writes \((A + B)^2 = A^2 + 2AB + B^2\) for two square matrices. Explain what is wrong, and write the correct expansion.

    • \((A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2.\)
    • The student has treated \(AB\) and \(BA\) as equal. In general \(AB \ne BA\), so they cannot be combined into \(2AB\).
    • \(\text{Correct: } (A + B)^2 = A^2 + AB + BA + B^2.\)

jmaths.xyz · AI HL · AHL 5.2 Matrix multiplication