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

Matrix multiplication Problem set B Worked solutions

Skills

  1. Work out each product, or say why it does not exist.

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

      • Each entry is a row of the first matrix times a column of the second:
      • \(\begin{pmatrix}(2)(1) + (-1)(-2) & (2)(0) + (-1)(5) \\ (4)(1) + (3)(-2) & (4)(0) + (3)(5)\end{pmatrix} = \begin{pmatrix}4 & -5 \\ -2 & 15\end{pmatrix}\)

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

      • Each entry is a row of the first matrix times a column of the second:
      • \(\begin{pmatrix}(2)(3) + (-1)(1) + (4)(-2)\end{pmatrix} = \begin{pmatrix}-3\end{pmatrix}\)

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

      • Each entry is a row of the first matrix times a column of the second:
      • \(\begin{pmatrix}(3)(1) & (3)(4) & (3)(-2) \\ (-1)(1) & (-1)(4) & (-1)(-2) \\ (2)(1) & (2)(4) & (2)(-2)\end{pmatrix} = \begin{pmatrix}3 & 12 & -6 \\ -1 & -4 & 2 \\ 2 & 8 & -4\end{pmatrix}\)

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

      • Each entry is a row of the first matrix times a column of the second:
      • \(\begin{pmatrix}(1)(2) + (-2)(-1) + (3)(3) & (1)(1) + (-2)(0) + (3)(5) \\ (0)(2) + (4)(-1) + (-1)(3) & (0)(1) + (4)(0) + (-1)(5)\end{pmatrix} = \begin{pmatrix}13 & 16 \\ -7 & -5\end{pmatrix}\)

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

      • The first matrix has 2 columns and the second has 3 rows. These must be equal, so the product does not exist.

    6. (f) \(\begin{pmatrix}3 & 5 \\ 7 & 2\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\)

      • Each entry is a row of the first matrix times a column of the second:
      • \(\begin{pmatrix}(3)(0) + (5)(1) & (3)(1) + (5)(0) \\ (7)(0) + (2)(1) & (7)(1) + (2)(0)\end{pmatrix} = \begin{pmatrix}5 & 3 \\ 2 & 7\end{pmatrix}\)

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

    1. (a) \(AB\)

      • A is 4 × 3 and B is 3 × 2. The inner numbers match (3), so the product is 4 × 2.

    2. (b) \(BA\)

      • B is 3 × 2 and A is 4 × 3. The inner numbers 2 and 4 are different, so it does not exist.

    3. (c) \(BC\)

      • B is 3 × 2 and C is 2 × 4. The inner numbers match (2), so the product is 3 × 4.

    4. (d) \(CA\)

      • C is 2 × 4 and A is 4 × 3. The inner numbers match (4), so the product is 2 × 3.

    5. (e) \(AC\)

      • A is 4 × 3 and C is 2 × 4. The inner numbers 3 and 2 are different, so it does not exist.

    6. (f) \(CB\)

      • C is 2 × 4 and B is 3 × 2. The inner numbers 4 and 3 are different, so it does not exist.

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

    1. (a) \(AB\)

      • \(AB = \begin{pmatrix}(3)(2) + (1)(4) & (3)(-1) + (1)(3) \\ (-2)(2) + (0)(4) & (-2)(-1) + (0)(3)\end{pmatrix} = \begin{pmatrix}10 & 0 \\ -4 & 2\end{pmatrix}\)

    2. (b) \(BA\)

      • \(BA = \begin{pmatrix}(2)(3) + (-1)(-2) & (2)(1) + (-1)(0) \\ (4)(3) + (3)(-2) & (4)(1) + (3)(0)\end{pmatrix} = \begin{pmatrix}8 & 2 \\ 6 & 4\end{pmatrix}\)
      • The results differ, so \(AB \ne BA\).

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

      • \(A^2 = \begin{pmatrix}(3)(3) + (1)(-2) & (3)(1) + (1)(0) \\ (-2)(3) + (0)(-2) & (-2)(1) + (0)(0)\end{pmatrix} = \begin{pmatrix}7 & 3 \\ -6 & -2\end{pmatrix}\)

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

      • \(A^2 - 2A + 3I = \begin{pmatrix}7 & 3 \\ -6 & -2\end{pmatrix} - \begin{pmatrix}6 & 2 \\ -4 & 0\end{pmatrix} + \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\)
      • \(= \begin{pmatrix}4 & 1 \\ -2 & 1\end{pmatrix}\)

  4. Find the unknowns.

    1. (a) \(\begin{pmatrix}x & 3 \\ 2 & y\end{pmatrix}\begin{pmatrix}2 \\ -1\end{pmatrix} = \begin{pmatrix}7 \\ 5\end{pmatrix}\)

      • Multiply out: \(\begin{pmatrix}2x - 3 \\ 4 - y\end{pmatrix} = \begin{pmatrix}7 \\ 5\end{pmatrix}\).
      • \(x = 5,\; y = -1\)

    2. (b) \(\begin{pmatrix}3 & p \\ 2 & 1\end{pmatrix}\begin{pmatrix}2 & 1 \\ q & 4\end{pmatrix} = \begin{pmatrix}0 & 11 \\ 1 & 6\end{pmatrix}\)

      • Multiply out: \(\begin{pmatrix}6 + pq & 3 + 4p \\ 4 + q & 6\end{pmatrix} = \begin{pmatrix}0 & 11 \\ 1 & 6\end{pmatrix}\).
      • Top right: \(3 + 4p = 11\) so \(p = 2\). Bottom left: \(4 + q = 1\) so \(q = -3\).
      • Check the top left: \(6 + (2)(-3) = 0\).

In context

  1. A garden centre sells tulip, rose and lily bulbs by the dozen. The prices in euros at its two branches, N and S, are \(P = \begin{pmatrix}3.2 & 3.5 \\ 5 & 4.8 \\ 2.4 & 2.6\end{pmatrix}\) (rows: tulips, roses, lilies; columns: N, S). (a) Branch N raises all its prices by 5% and branch S cuts all its prices by 2%. Write this as a matrix product and find the new prices. (b) Instead, tulips go up 8%, roses stay the same and lilies go down 10%. Write this as a matrix product and find the new prices.

    • (a) Each column of prices is multiplied by its own factor, so \(P\) is multiplied on the right by a diagonal matrix: \(P\begin{pmatrix}1.05 & 0 \\ 0 & 0.98\end{pmatrix} = \begin{pmatrix}3.36 & 3.43 \\ 5.25 & 4.704 \\ 2.52 & 2.548\end{pmatrix}\). To the cent: \(\begin{pmatrix}3.36 & 3.43 \\ 5.25 & 4.70 \\ 2.52 & 2.55\end{pmatrix}\).
    • (b) Each row of prices is multiplied by its own factor, so the diagonal matrix goes on the left: \(\begin{pmatrix}1.08 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.9\end{pmatrix}P = \begin{pmatrix}3.456 & 3.78 \\ 5 & 4.8 \\ 2.16 & 2.34\end{pmatrix}\). To the cent: \(\begin{pmatrix}3.46 & 3.78 \\ 5.00 & 4.80 \\ 2.16 & 2.34\end{pmatrix}\).

  2. A museum runs three exhibitions. The numbers of adult and child tickets sold are \(\begin{pmatrix}110 & 35 \\ 90 & 60 \\ 75 & 40\end{pmatrix}\). An adult ticket costs €11.50 and a child ticket costs €7. Use a matrix product to find the takings for each exhibition.

    • Put the prices in a column and multiply: \(\begin{pmatrix}110 & 35 \\ 90 & 60 \\ 75 & 40\end{pmatrix}\begin{pmatrix}11.5 \\ 7\end{pmatrix}\).
    • \(= \begin{pmatrix}(110)(11.5) + (35)(7) \\ (90)(11.5) + (60)(7) \\ (75)(11.5) + (40)(7)\end{pmatrix}\)
    • \(= \begin{pmatrix}1510 \\ 1455 \\ 1142.5\end{pmatrix}\)

  3. In a survey, the proportions of Grade 11 and Grade 12 students choosing each of three after-school clubs were \(P = \begin{pmatrix}0.3 & 0.2 \\ 0.45 & 0.5 \\ 0.25 & 0.3\end{pmatrix}\) (columns: Grade 11, Grade 12). Two campuses have \(N = \begin{pmatrix}160 & 240 \\ 200 & 120\end{pmatrix}\) students (rows: Grade 11, Grade 12; columns: campus 1, campus 2). Assuming the proportions hold, find the matrix giving the expected number of students in each club on each campus, and the total number of places needed in the second club.

    • \(PN = \begin{pmatrix}0.3 & 0.2 \\ 0.45 & 0.5 \\ 0.25 & 0.3\end{pmatrix}\begin{pmatrix}160 & 240 \\ 200 & 120\end{pmatrix}\)
    • \(= \begin{pmatrix}88 & 96 \\ 172 & 168 \\ 100 & 96\end{pmatrix}\)
    • The second club is the second row: 172 + 168 = 340.

  4. A bakery makes two cakes, lemon and chocolate. In one day it makes \(\begin{pmatrix}30 & 20\end{pmatrix}\) (lemon, chocolate). The packs of flour, sugar and butter each cake needs are \(\begin{pmatrix}3 & 2 & 1 \\ 2 & 2 & 3\end{pmatrix}\) (rows: lemon, chocolate). Two wholesalers charge, in euros per pack, \(\begin{pmatrix}0.9 & 0.85 \\ 1.2 & 1.25 \\ 1.5 & 1.4\end{pmatrix}\) (rows: flour, sugar, butter; columns: wholesaler 1, wholesaler 2). Find the cost of the day's packs from each wholesaler, and say which is cheaper and by how much.

    • \(\text{Orders: } (1 \times 2)(2 \times 3)(3 \times 2) \text{ gives } 1 \times 2.\)
    • \(OD = \begin{pmatrix}130 & 100 & 90\end{pmatrix}\)
    • \(ODP = \begin{pmatrix}372 & 361.5\end{pmatrix}\)
    • Wholesaler 1 costs €372.00 and wholesaler 2 costs €361.50, so wholesaler 2 is cheaper by €10.50.

Stretch

  1. Find the value of \(k\) for which \(\begin{pmatrix}2 & 1 \\ 0 & 4\end{pmatrix}\) and \(\begin{pmatrix}k & 6 \\ 0 & 3\end{pmatrix}\) commute.

    • Both products are upper triangular, so only the top-right element can differ.
    • In \(AB\) it is \((2)(6) + (1)(3) = 15\). In \(BA\) it is \((k)(1) + (6)(4) = k + 24\).
    • \(15 = k + 24\) gives \(k = -9\).

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

    • \(A^2 = \begin{pmatrix}(1)(1) + (3)(0) & (1)(3) + (3)(1) \\ (0)(1) + (1)(0) & (0)(3) + (1)(1)\end{pmatrix} = \begin{pmatrix}1 & 6 \\ 0 & 1\end{pmatrix}\)
    • \(A^3 = A^2A = \begin{pmatrix}(1)(1) + (6)(0) & (1)(3) + (6)(1) \\ (0)(1) + (1)(0) & (0)(3) + (1)(1)\end{pmatrix} = \begin{pmatrix}1 & 9 \\ 0 & 1\end{pmatrix}\)
    • The top-right element goes up by 3 each time, so \(A^n = \begin{pmatrix}1 & 3n \\ 0 & 1\end{pmatrix}\).

  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.

    • Expand carefully, keeping the order of each product: \((A - B)(A - B) = A^2 - AB - BA + B^2\).
    • The student has treated \(BA\) as if it were \(AB\). Matrix multiplication is not commutative, so \(-AB - BA\) cannot be written as \(-2AB\) unless \(AB = BA\).

jmaths.xyz · AI HL · AHL 5.2 Matrix multiplication