Matrix arithmetic Problem set B One question at a time

State the order of each matrix.

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

   • It has 4 rows and 2 columns, so the order is 4 × 2 (rows × columns).

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

   • It has 1 row and 5 columns, so the order is 1 × 5 (rows × columns).

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

   • It has 3 rows and 3 columns, so the order is 3 × 3 (rows × columns).

Find \(M\) in each case.

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

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

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

   • \(2M = \begin{pmatrix}-1 & -4 \\ 3 & 2\end{pmatrix} + \begin{pmatrix}3 & -1 \\ -2 & 4\end{pmatrix} = \begin{pmatrix}2 & -5 \\ 1 & 6\end{pmatrix}\)
   • \(M = \tfrac{1}{2}\begin{pmatrix}2 & -5 \\ 1 & 6\end{pmatrix} = \begin{pmatrix}1 & -2.5 \\ 0.5 & 3\end{pmatrix}\)

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

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

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

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

   • \(B - A = \begin{pmatrix}-3 & 6 \\ 4 & -5\end{pmatrix} - \begin{pmatrix}3 & -2 \\ 1 & 4\end{pmatrix} = \begin{pmatrix}-6 & 8 \\ 3 & -9\end{pmatrix}\)

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

   • Matrices can only be added when they have the same order. \(A\) is 2 × 2 and \(C\) is 2 × 3, so it cannot be done.

4. (d) \(4A\)

   • \(4A = 4\begin{pmatrix}3 & -2 \\ 1 & 4\end{pmatrix} = \begin{pmatrix}12 & -8 \\ 4 & 16\end{pmatrix}\)

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

   • \(3A - 2B = \begin{pmatrix}9 & -6 \\ 3 & 12\end{pmatrix} - \begin{pmatrix}-6 & 12 \\ 8 & -10\end{pmatrix} = \begin{pmatrix}15 & -18 \\ -5 & 22\end{pmatrix}\)

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

   • \(\tfrac{1}{2}B = \tfrac{1}{2}\begin{pmatrix}-3 & 6 \\ 4 & -5\end{pmatrix} = \begin{pmatrix}-1.5 & 3 \\ 2 & -2.5\end{pmatrix}\)

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

1. (a) \(\begin{pmatrix}x-2 & 0 \\ 7 & 2y\end{pmatrix} = \begin{pmatrix}5 & 0 \\ 7 & y+6\end{pmatrix}\)

   • Equal matrices have equal entries: \(x - 2 = 5\) and \(2y = y + 6\).
   • \(x = 7,\; y = 6\)

2. (b) \(\begin{pmatrix}x^2 & -1 \\ y & 4\end{pmatrix} = \begin{pmatrix}36 & -1 \\ 2x & 4\end{pmatrix}\)

   • \(x^2 = 36\) so \(x = 6\) or \(x = -6\), and \(y = 2x\).
   • \(x = 6,\; y = 12\) or \(x = -6,\; y = -12\)

A bike shop stocks three frame sizes of two kinds of bicycle. The stock at the start of the week is \(S = \begin{pmatrix}12 & 8 & 15 \\ 9 & 14 & 10\end{pmatrix}\). During the week it sells \(D = \begin{pmatrix}5 & 3 & 9 \\ 4 & 6 & 2\end{pmatrix}\) and takes a delivery of \(R = \begin{pmatrix}8 & 0 & 6 \\ 0 & 5 & 10\end{pmatrix}\). Write a matrix expression for the stock at the end of the week, and work it out.

• Stock at the end = start - sold + delivered.
• \(S - D + R = \begin{pmatrix}12 & 8 & 15 \\ 9 & 14 & 10\end{pmatrix} - \begin{pmatrix}5 & 3 & 9 \\ 4 & 6 & 2\end{pmatrix} + \begin{pmatrix}8 & 0 & 6 \\ 0 & 5 & 10\end{pmatrix}\)
• \(= \begin{pmatrix}7 & 5 & 6 \\ 5 & 8 & 8\end{pmatrix} + \begin{pmatrix}8 & 0 & 6 \\ 0 & 5 & 10\end{pmatrix} = \begin{pmatrix}15 & 5 & 12 \\ 5 & 13 & 18\end{pmatrix}\)

The monthly fees, in dollars, for a single and a family membership at two gyms are \(\begin{pmatrix}45 & 80 \\ 62 & 95\end{pmatrix}\). All fees rise by 8%. Find the matrix of new fees.

• A rise of 8% multiplies every fee by 1.08.
• \(1.08\begin{pmatrix}45 & 80 \\ 62 & 95\end{pmatrix} = \begin{pmatrix}48.6 & 86.4 \\ 66.96 & 102.6\end{pmatrix}\)

Two quiz teams get the following numbers of correct answers in the three rounds of the autumn league, \(\begin{pmatrix}7 & 5 & 9 \\ 4 & 8 & 6\end{pmatrix}\), and of the spring league, \(\begin{pmatrix}6 & 9 & 5 \\ 8 & 3 & 7\end{pmatrix}\). Find the matrix of total correct answers. Which team got more correct answers altogether?

• Add the matrices: \(\begin{pmatrix}7 & 5 & 9 \\ 4 & 8 & 6\end{pmatrix} + \begin{pmatrix}6 & 9 & 5 \\ 8 & 3 & 7\end{pmatrix} = \begin{pmatrix}13 & 14 & 14 \\ 12 & 11 & 13\end{pmatrix}\)
• Row totals: 13 + 14 + 14 = 41 for the first team and 12 + 11 + 13 = 36 for the second.
• The first team got more.

Find \(p\) and \(q\) such that \(p\begin{pmatrix}3 & 1 \\ 0 & 2\end{pmatrix} + q\begin{pmatrix}1 & 2 \\ -3 & 1\end{pmatrix} = \begin{pmatrix}9 & 8 \\ -9 & 7\end{pmatrix}\).

• Compare entries: \(3p + q = 9\), \(p + 2q = 8\), \(-3q = -9\), \(2p + q = 7\).
• \(-3q = -9\) gives \(q = 3\), then \(3p + 3 = 9\) gives \(p = 2\).
• Check: \(p + 2q = 2 + 6 = 8\) and \(2p + q = 7\).

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

• \(A + sB = \begin{pmatrix}a + 2s & 3 + sb \\ -2 + s & 1 + 3s\end{pmatrix} = \begin{pmatrix}t & 0 \\ 0 & t\end{pmatrix}\)
• From the bottom left, \(-2 + s = 0\) so \(s = 2\). Then \(t = 1 + 3s = 7\).
• From the top right, \(3 + 2b = 0\) so \(b = -1.5\). From the top left, \(a + 4 = 7\) so \(a = 3\).