Matrix arithmetic Problem set A Worked solutions

State the order of each matrix.

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

   • 3 rows and 2 columns, so the order is \(3 \times 2\).

2. (b) \(\begin{pmatrix}7 & 0 & -2 & 1\end{pmatrix}\)

   • 1 row and 4 columns, so the order is \(1 \times 4\).

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

   • 3 rows and 3 columns, so the order is \(3 \times 3\).

Find \(M\) in each case.

1. (a) \(M + \begin{pmatrix}4 & -1 \\ 2 & 0\end{pmatrix} = \begin{pmatrix}1 & 5 \\ -3 & 7\end{pmatrix}\)

   • \(M = \begin{pmatrix}1 & 5 \\ -3 & 7\end{pmatrix} - \begin{pmatrix}4 & -1 \\ 2 & 0\end{pmatrix}\)
   • \(M = \begin{pmatrix}-3 & 6 \\ -5 & 7\end{pmatrix}\)

2. (b) \(2M - \begin{pmatrix}2 & -5 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}-1 & 4 \\ 6 & -2\end{pmatrix}\)

   • \(2M = \begin{pmatrix}-1 & 4 \\ 6 & -2\end{pmatrix} + \begin{pmatrix}2 & -5 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}1 & -1 \\ 6 & 1\end{pmatrix}\)
   • \(M = \tfrac{1}{2}\begin{pmatrix}1 & -1 \\ 6 & 1\end{pmatrix} = \begin{pmatrix}0.5 & -0.5 \\ 3 & 0.5\end{pmatrix}\)

\(A = \begin{pmatrix}2 & -5 \\ 0 & 3\end{pmatrix},\; B = \begin{pmatrix}-1 & 4 \\ 6 & -2\end{pmatrix},\; C = \begin{pmatrix}1 & 0 & 2 \\ -3 & 5 & 1\end{pmatrix}\). Find each of these, or say why it cannot be done.

1. (a) \(A + B\)

   • \(A + B = \begin{pmatrix}1 & -1 \\ 6 & 1\end{pmatrix}\)

2. (b) \(B - A\)

   • \(B - A = \begin{pmatrix}-3 & 9 \\ 6 & -5\end{pmatrix}\)

3. (c) \(A + C\)

   • \(A\) is \(2 \times 2\) and \(C\) is \(2 \times 3\). Matrices can be added only when they have the same order.
   • Cannot be done.

4. (d) \(3A\)

   • \(3A = \begin{pmatrix}6 & -15 \\ 0 & 9\end{pmatrix}\)

5. (e) \(2A - 3B\)

   • \(2A = \begin{pmatrix}4 & -10 \\ 0 & 6\end{pmatrix},\quad 3B = \begin{pmatrix}-3 & 12 \\ 18 & -6\end{pmatrix}\)
   • \(2A - 3B = \begin{pmatrix}7 & -22 \\ -18 & 12\end{pmatrix}\)

6. (f) \(\tfrac{1}{2}B\)

   • \(\tfrac{1}{2}B = \begin{pmatrix}-0.5 & 2 \\ 3 & -1\end{pmatrix}\)

Find \(x\) and \(y\).

1. (a) \(\begin{pmatrix}x+3 & -2 \\ 4 & 2y\end{pmatrix} = \begin{pmatrix}7 & -2 \\ 4 & y+5\end{pmatrix}\)

   • \(\text{Equal matrices have equal entries: } x + 3 = 7 \text{ and } 2y = y + 5.\)
   • \(x = 4,\; y = 5\)

2. (b) \(\begin{pmatrix}x^2 & 1 \\ y & 6\end{pmatrix} = \begin{pmatrix}16 & 1 \\ -x & 6\end{pmatrix}\)

   • \(\text{Top-left: } x^2 = 16 \text{, so } x = 4 \text{ or } x = -4.\)
   • \(\text{Bottom-left: } y = -x.\)
   • \(x = 4,\; y = -4 \quad \text{or} \quad x = -4,\; y = 4\)

A shop stocks three sizes of two kinds of trainer. The stock at the start of the week is \(S = \begin{pmatrix}14 & 9 & 20 \\ 8 & 11 & 16\end{pmatrix}\). During the week it sells \(D = \begin{pmatrix}6 & 4 & 11 \\ 3 & 7 & 5\end{pmatrix}\) and takes a delivery of \(R = \begin{pmatrix}10 & 0 & 12 \\ 5 & 10 & 0\end{pmatrix}\). Write a matrix expression for the stock at the end of the week, and work it out.

• \(\text{End stock} = S - D + R\)
• \(= \begin{pmatrix}14 & 9 & 20 \\ 8 & 11 & 16\end{pmatrix} - \begin{pmatrix}6 & 4 & 11 \\ 3 & 7 & 5\end{pmatrix} + \begin{pmatrix}10 & 0 & 12 \\ 5 & 10 & 0\end{pmatrix}\)
• \(S - D = \begin{pmatrix}8 & 5 & 9 \\ 5 & 4 & 11\end{pmatrix}\)
• \(S - D + R = \begin{pmatrix}18 & 5 & 21 \\ 10 & 14 & 11\end{pmatrix}\)

The monthly rents, in euros, for one-bedroom and two-bedroom flats in two districts are \(\begin{pmatrix}850 & 1200 \\ 640 & 990\end{pmatrix}\). All rents rise by 6%. Find the matrix of new rents.

• A 6% rise means multiplying every entry by \(1.06\).
• \(1.06\begin{pmatrix}850 & 1200 \\ 640 & 990\end{pmatrix}\)
• \(= \begin{pmatrix}901 & 1272 \\ 678.4 & 1049.4\end{pmatrix}\)

Two swimming clubs win gold, silver and bronze medals at the autumn gala, \(\begin{pmatrix}4 & 6 & 3 \\ 7 & 2 & 5\end{pmatrix}\), and at the spring gala, \(\begin{pmatrix}5 & 3 & 8 \\ 6 & 6 & 1\end{pmatrix}\). Find the matrix of total medals. Which club won more medals altogether?

• \(\text{Total} = \begin{pmatrix}4 & 6 & 3 \\ 7 & 2 & 5\end{pmatrix} + \begin{pmatrix}5 & 3 & 8 \\ 6 & 6 & 1\end{pmatrix} = \begin{pmatrix}9 & 9 & 11 \\ 13 & 8 & 6\end{pmatrix}\)
• \(\text{Club 1: } 9 + 9 + 11 = 29,\quad \text{club 2: } 13 + 8 + 6 = 27\)
• The first club won more medals: 29 against 27.

Find \(p\) and \(q\) such that \(p\begin{pmatrix}2 & 1 \\ 0 & 3\end{pmatrix} + q\begin{pmatrix}1 & -1 \\ 4 & 0\end{pmatrix} = \begin{pmatrix}7 & 2 \\ 4 & 9\end{pmatrix}\).

• \(\text{Equate entries: } 2p + q = 7,\quad p - q = 2,\quad 4q = 4,\quad 3p = 9.\)
• \(4q = 4 \text{ gives } q = 1; \; 3p = 9 \text{ gives } p = 3.\)
• \(\text{Check: } 2(3) + 1 = 7 \text{ and } 3 - 1 = 2.\)
• \(p = 3,\; q = 1\)

\(A = \begin{pmatrix}a & 2 \\ -1 & 3\end{pmatrix}\) and \(B = \begin{pmatrix}1 & b \\ 2 & 4\end{pmatrix}\). Given that \(A + sB = tI\), find \(a, b, s\) and \(t\).

• \(A + sB = \begin{pmatrix}a + s & 2 + bs \\ -1 + 2s & 3 + 4s\end{pmatrix}\)
• \(tI = \begin{pmatrix}t & 0 \\ 0 & t\end{pmatrix}\)
• \(\text{Bottom-left: } -1 + 2s = 0, \text{ so } s = \tfrac{1}{2}.\)
• \(\text{Top-right: } 2 + sb = 0, \text{ so } b = -4.\)
• \(\text{Bottom-right: } t = 3 + 4s = 5.\)
• \(\text{Top-left: } a + s = t, \text{ so } a = 5 - \tfrac{1}{2} = 4.5.\)