MATRICES

Chapter 3

Matrices

Big idea: a whole table of numbers can be treated as one object, and then added, scaled, multiplied, flipped and undone, just like a single number.

A matrix is a table of numbers that is treated as one single object. Tables like this are everywhere:

• A spreadsheet: every number has a row and a column.
• A photo on a screen: the screen is a grid of tiny dots (pixels), and every dot has a number for its brightness.
• A game scoreboard: players go down the side and rounds go across the top.
• A weather chart: cities go down the side and days go across the top.

For example, here is a tiny $3 \times 3$ picture. Each number tells how bright a pixel is, from 0 (black) to 255 (white):

$$\begin{bmatrix} 0 & 60 & 120 \\ 60 & 120 & 200 \\ 120 & 200 & 255 \end{bmatrix}$$

A computer stores and changes the whole picture by working with this one table. To make the picture brighter, it changes every number in a single step. This is the power of a matrix: we can work with a whole table at once.

Matrices are used to:

• store and combine large tables of data (sales, scores, measurements);
• move, turn and resize shapes in computer graphics and games;
• solve many equations at the same time;
• send secret messages in cryptography;
• describe how a system changes from one step to the next, in science and economics.

Think about it: which table of numbers do you use most often in daily life? How many rows and columns does it have?

Three students, A, B and C, take two tests. Their marks are shown in this table:

If we agree that rows are students (in the order A, B, C) and columns are tests (in the order 1, 2), we do not need the labels. We can keep only the numbers and put them inside brackets:

$$M = \begin{bmatrix} 72 & 65 \\ 58 & 81 \\ 90 & 77 \end{bmatrix}$$

This arrangement is a matrix.

Definition. A matrix is a rectangular arrangement of numbers (or expressions) in rows and columns, written inside brackets. Each number in it is called an element or entry. We name matrices with capital letters such as $A$, $B$ and $M$.

• A row is a horizontal line of entries. Rows are counted from the top.
• A column is a vertical line of entries. Columns are counted from the left.

In $M$ above, row 1 is $[72 \;\; 65]$ and column 2 is $\begin{bmatrix} 65 \\ 81 \\ 77 \end{bmatrix}$.

$$A = \begin{bmatrix} 2 & -1 & 4 \\ 0 & 3 & 7 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 & 0 \\ -2 & 9 \\ 1 & 6 \end{bmatrix}, \qquad C = \begin{bmatrix} 1.5 & -3 & 8 & 0 \end{bmatrix}$$

$A$ has 2 rows and 3 columns. $B$ has 3 rows and 2 columns. $C$ has 1 row and 4 columns. Entries can be positive, negative, zero, fractions or decimals.

1. In matrix $M$ above, which student got the highest mark in Test 1, and what was it?
2. How many rows and columns does $\begin{bmatrix} 4 & 1 & 0 \\ 7 & 2 & 9 \\ 3 & 5 & 8 \\ 6 & 0 & 1 \end{bmatrix}$ have?
3. Write the marks of the three students in Test 2 as a column.

Answers

1. Student C, with 90 (the largest entry in column 1).
2. 4 rows and 3 columns.
3. $\begin{bmatrix} 65 \\ 81 \\ 77 \end{bmatrix}$

The order of a matrix tells us its size. A matrix with $m$ rows and $n$ columns has order $m \times n$, read "$m$ by $n$". The number of rows always comes first.

For example, $\begin{bmatrix} 3 & 1 & 4 \\ 1 & 5 & 9 \end{bmatrix}$ has order $2 \times 3$ and has $2 \times 3 = 6$ entries.

We write $a_{ij}$ for the entry in row $i$ and column $j$. A matrix of order $m \times n$ can then be written as $A = [a_{ij}]_{m \times n}$:

$$A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}$$

Example 1. Construct a $3 \times 2$ matrix $A = [a_{ij}]$ whose entries are given by $a_{ij} = 2i - j$.

Solution. A $3 \times 2$ matrix has entries $a_{11}, a_{12}, a_{21}, a_{22}, a_{31}, a_{32}$. Put in the row number $i$ and the column number $j$ each time:

$$\begin{aligned} a_{11} &= 2(1) - 1 = 1, & a_{12} &= 2(1) - 2 = 0, \\ a_{21} &= 2(2) - 1 = 3, & a_{22} &= 2(2) - 2 = 2, \\ a_{31} &= 2(3) - 1 = 5, & a_{32} &= 2(3) - 2 = 4. \end{aligned}$$

So $A = \begin{bmatrix} 1 & 0 \\ 3 & 2 \\ 5 & 4 \end{bmatrix}$.

Example 2. A matrix has 12 entries. What orders can it have?

Solution. We need $m \times n = 12$ with $m$ and $n$ natural numbers. The pairs are $(1, 12)$, $(2, 6)$, $(3, 4)$, $(4, 3)$, $(6, 2)$, $(12, 1)$. So the possible orders are $1 \times 12$, $2 \times 6$, $3 \times 4$, $4 \times 3$, $6 \times 2$, $12 \times 1$.

Example 3. A matrix can store the corners of a shape. The corners of a four-sided figure PQRS are $P(0, 0)$, $Q(4, 1)$, $R(5, 4)$ and $S(1, 3)$. Write them as the columns of a matrix, so that the first row holds the $x$-coordinates and the second row holds the $y$-coordinates.

Solution. $X = \begin{bmatrix} 0 & 4 & 5 & 1 \\ 0 & 1 & 4 & 3 \end{bmatrix}$ This matrix has order $2 \times 4$.

Computer graphics programs store every shape this way. To move or turn the shape, they change the numbers in this one matrix.

1. What is the order of $\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \\ 7 & 8 \end{bmatrix}$, and how many entries does it have?
2. A matrix has 15 entries. List all possible orders.
3. Write the $2 \times 2$ matrix with $a_{ij} = i + j$.
4. In a matrix of order $4 \times 5$, how many entries lie in row 3?

Answers

1. Order $4 \times 2$; 8 entries.
2. $1 \times 15$, $3 \times 5$, $5 \times 3$, $15 \times 1$.
3. $\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}$
4. 5 entries (one for each column).

Some matrices have a special shape or a special pattern, and each has its own name. Knowing the names helps us spot shortcuts later.

A matrix with only one column is a column matrix. Its order is $m \times 1$.

Example: the prices of four items listed one below the other form the column matrix $\begin{bmatrix} 250 \\ 40 \\ 75 \\ 190 \end{bmatrix}$, of order $4 \times 1$.

A matrix with only one row is a row matrix. Its order is $1 \times n$.

Example: one player's scores in five rounds, written side by side, form the row matrix $\begin{bmatrix} 6 & 0 & 9 & 4 & 7 \end{bmatrix}$, of order $1 \times 5$.

A matrix with the same number of rows and columns is a square matrix. If it has $n$ rows and $n$ columns, we say it is a square matrix of order $n$.

Example: $S = \begin{bmatrix} 4 & -2 & 7 \\ 0 & 5 & 1 \\ 3 & 8 & -6 \end{bmatrix}$ is a square matrix of order 3. A tic-tac-toe board and a sudoku grid are square.

The entries $a_{11}, a_{22}, \dots, a_{nn}$ of a square matrix form its diagonal. In $S$ the diagonal entries are 4, 5 and −6.

A square matrix in which every entry off the diagonal is 0 is a diagonal matrix. In symbols, $a_{ij} = 0$ whenever $i \neq j$.

Example: $D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 6 \end{bmatrix}$ is a diagonal matrix of order 3.

A diagonal matrix whose diagonal entries are all equal is a scalar matrix. So $a_{ij} = 0$ for $i \neq j$ and $a_{ii} = k$ for one fixed number $k$.

Example: $\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix}$ is a scalar matrix with $k = 7$.

A scalar matrix with $k = 1$, that is, with 1 on the diagonal and 0 elsewhere, is an identity matrix. We write $I_n$ for the identity matrix of order $n$, or simply $I$ when the order is clear.

$$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \qquad I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Every identity matrix is a scalar matrix, but a scalar matrix is an identity matrix only when $k = 1$.

A matrix whose entries are all 0 is a zero matrix (or null matrix). We write $O$ for it; its order is understood from the situation.

$$O_{2 \times 2} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \qquad O_{1 \times 3} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$$

1. Name the type of each matrix: (a) $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ (b) $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ (c) $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ (d) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (e) $\begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix}$.
2. Is every diagonal matrix a square matrix? Is every square matrix a diagonal matrix?

Answers

1. (a) zero matrix (b) column matrix (c) scalar matrix (and diagonal, square) (d) identity matrix (and scalar, diagonal, square) (e) square matrix only (it has a non-zero entry off the diagonal).
2. Yes, every diagonal matrix is square by definition. No, a square matrix such as the one in (e) is not diagonal.

Two matrices $A$ and $B$ are equal (we write $A = B$) when both of these are true:

1. they have the same order, and
2. every entry of $A$ equals the entry of $B$ in the same position, that is, $a_{ij} = b_{ij}$ for all $i$ and $j$.

For example, $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, but $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$, because the entries in row 1 are swapped. Also, $[1 \;\; 2] \neq \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ because their orders differ.

1. Pair up the entries that sit in the same position.
2. Write one small equation for each pair.
3. Solve each equation.

Example 1. Find $x$ and $y$ if $\begin{bmatrix} x + 2 & 5 \\ 3 & y - 1 \end{bmatrix} = \begin{bmatrix} 7 & 5 \\ 3 & 4 \end{bmatrix}$.

Solution. Compare the entries in the same positions. Row 1, column 1 gives $x + 2 = 7$, so $x = 5$. Row 2, column 2 gives $y - 1 = 4$, so $y = 5$. The other two pairs ($5 = 5$ and $3 = 3$) already match. Hence $x = 5$ and $y = 5$.

Example 2. Find $x$, $y$ and $z$ if $\begin{bmatrix} 2x + y & 4 \\ z & x - y \end{bmatrix} = \begin{bmatrix} 7 & 4 \\ -1 & -1 \end{bmatrix}$.

Solution. Comparing entries gives three equations: $2x + y = 7$, $z = -1$ and $x - y = -1$. Adding the first and third equations gives $3x = 6$, so $x = 2$. Then $y = x + 1 = 3$. Check: $2(2) + 3 = 7$ and $2 - 3 = -1$. Hence $x = 2$, $y = 3$, $z = -1$.

1. Are $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ equal? Why or why not?
2. Find $a$ and $b$ if $\begin{bmatrix} a + b & 3 \\ 0 & a - b \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 0 & 2 \end{bmatrix}$.

Answers

1. No. The orders are different ($2 \times 2$ and $2 \times 3$).
2. $a + b = 8$ and $a - b = 2$; adding gives $2a = 10$, so $a = 5$ and $b = 3$.

1. For $A = \begin{bmatrix} 3 & -2 & 8 & 1 \\ 0 & 5 & -4 & 9 \\ 7 & 2 & 6 & -1 \end{bmatrix}$ write: (a) the order of $A$, (b) the number of entries, (c) the entries $a_{13}$, $a_{22}$ and $a_{34}$.
2. List all possible orders of a matrix with (a) 18 entries (b) 7 entries.
3. Construct the $2 \times 3$ matrix with $a_{ij} = ij + 1$.
4. Construct the $3 \times 2$ matrix with $a_{ij} = (i - j)^2$.
5. Find $x$ and $y$ if $\begin{bmatrix} x + y & 2 \\ 4 & x - y \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 4 & 3 \end{bmatrix}$.
6. Find $p$, $q$ and $r$ if $\begin{bmatrix} p & q + 1 \\ 2r & 5 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 10 & 5 \end{bmatrix}$.
7. Choose the correct answer. A matrix $[a_{ij}]_{m \times n}$ is a square matrix if (A) $m < n$ (B) $m > n$ (C) $m = n$ (D) none of these.
8. How many different $2 \times 2$ matrices can be made if every entry must be 0 or 1?

Answers

1. (a) $3 \times 4$; (b) 12; (c) $a_{13} = 8$, $a_{22} = 5$, $a_{34} = -1$.
2. (a) $1 \times 18$, $2 \times 9$, $3 \times 6$, $6 \times 3$, $9 \times 2$, $18 \times 1$; (b) $1 \times 7$, $7 \times 1$.
3. $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 5 & 7 \end{bmatrix}$
4. $\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 4 & 1 \end{bmatrix}$
5. $x = 6$, $y = 3$.
6. $p = 4$, $q = 7$, $r = 5$.
7. (C).
8. $2^4 = 16$ matrices (each of the 4 entries has 2 choices).

Just as we add, subtract and multiply numbers, we can combine matrices. In this section we learn four operations:

• Addition of matrices
• Multiplication of a matrix by a scalar (a scalar is an ordinary number)
• Subtraction of matrices
• Multiplication of two matrices

A warehouse supplies laptops, tablets and phones to two shops, Shop 1 and Shop 2. The tables show the number of devices delivered in Week 1 and in Week 2 (rows: shops; columns: laptops, tablets, phones):

$$\text{Week 1: } W_1 = \begin{bmatrix} 12 & 8 & 5 \\ 7 & 14 & 9 \end{bmatrix}, \qquad \text{Week 2: } W_2 = \begin{bmatrix} 6 & 10 & 3 \\ 9 & 4 & 11 \end{bmatrix}$$

To find the total delivered over the two weeks, we add the numbers that are in the same position:

$$W_1 + W_2 = \begin{bmatrix} 12 + 6 & 8 + 10 & 5 + 3 \\ 7 + 9 & 14 + 4 & 9 + 11 \end{bmatrix} = \begin{bmatrix} 18 & 18 & 8 \\ 16 & 18 & 20 \end{bmatrix}$$

So Shop 1 received 18, 18, 8 laptops, tablets and phones in total, and Shop 2 received 16, 18, 20.

Matrices of different orders cannot be added. A $2 \times 3$ table and a $3 \times 2$ table do not line up cell by cell, so their sum is not defined.

Example 1. Find $A + B$ if $A = \begin{bmatrix} 2 & -3 & 5 \\ 0 & 4 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 7 & 1 & -2 \\ -6 & 3 & 8 \end{bmatrix}$.

Solution. Both matrices have order $2 \times 3$, so $A + B$ is defined: $A + B = \begin{bmatrix} 2 + 7 & -3 + 1 & 5 + (-2) \\ 0 + (-6) & 4 + 3 & -1 + 8 \end{bmatrix} = \begin{bmatrix} 9 & -2 & 3 \\ -6 & 7 & 7 \end{bmatrix}$

Example 2. Can $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ be added?

Solution. No. The first has order $2 \times 2$ and the second has order $2 \times 3$. The orders are different, so the sum is not defined.

1. Find $\begin{bmatrix} 4 & 0 \\ -2 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ 2 & -3 \end{bmatrix}$.
2. Find $\begin{bmatrix} 1.5 & 2 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} 0.5 & -2 \\ 4 & 1 \end{bmatrix}$.
3. A $3 \times 4$ matrix is added to a $4 \times 3$ matrix. Is the sum defined?

Answers

1. $\begin{bmatrix} 5 & 5 \\ 0 & 4 \end{bmatrix}$
2. $\begin{bmatrix} 2 & 0 \\ 4 & 0 \end{bmatrix}$
3. No, the orders differ.

A scalar is an ordinary number. Suppose a recipe table $R$ lists the amount of each of three ingredients needed for two dishes when cooking for 2 people. To cook for 6 people we need 3 times as much of everything:

$$R = \begin{bmatrix} 2 & 1 & 4 \\ 3 & 0 & 2 \end{bmatrix}, \qquad 3R = \begin{bmatrix} 3 \cdot 2 & 3 \cdot 1 & 3 \cdot 4 \\ 3 \cdot 3 & 3 \cdot 0 & 3 \cdot 2 \end{bmatrix} = \begin{bmatrix} 6 & 3 & 12 \\ 9 & 0 & 6 \end{bmatrix}$$

We define $-A = (-1)A$. It is $A$ with the sign of every entry changed.

$$A = \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -3 & 1 \end{bmatrix}, \qquad -A = \begin{bmatrix} -5 & 2 \\ 0 & -4 \\ 3 & -1 \end{bmatrix}$$

For two matrices of the same order, $A - B$ means subtract entry by entry. This is the same as $A + (-B)$.

$$A - B = \begin{bmatrix} 5 - 1 & -2 - 3 \\ 0 - (-4) & 4 - 2 \\ -3 - 6 & 1 - (-5) \end{bmatrix} = \begin{bmatrix} 4 & -5 \\ 4 & 2 \\ -9 & 6 \end{bmatrix}$$

where $B = \begin{bmatrix} 1 & 3 \\ -4 & 2 \\ 6 & -5 \end{bmatrix}$. Real-life use: this week's sales minus last week's sales shows how each number has changed.

Example. With $A$ and $B$ as above, find $2A - B$.

Solution. First multiply, then subtract: $2A = \begin{bmatrix} 10 & -4 \\ 0 & 8 \\ -6 & 2 \end{bmatrix}, \qquad 2A - B = \begin{bmatrix} 10 - 1 & -4 - 3 \\ 0 - (-4) & 8 - 2 \\ -6 - 6 & 2 - (-5) \end{bmatrix} = \begin{bmatrix} 9 & -7 \\ 4 & 6 \\ -12 & 7 \end{bmatrix}$

1. Find $4\begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$.
2. If $P = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$, find $\frac{1}{2}P$.
3. Find $\begin{bmatrix} 9 & 5 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 5 \\ -1 & 3 \end{bmatrix}$.

Answers

1. $\begin{bmatrix} 4 & -8 \\ 0 & 12 \end{bmatrix}$
2. $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
3. $\begin{bmatrix} 5 & 0 \\ 3 & -2 \end{bmatrix}$

Matrix addition obeys the same friendly rules as addition of numbers. Take $A$, $B$ and $C$ of the same order.

• Commutative law: $A + B = B + A$. The order does not matter.
• Associative law: $(A + B) + C = A + (B + C)$. The grouping does not matter.
• Additive identity: $A + O = O + A = A$, where $O$ is the zero matrix of the same order.
• Additive inverse: $A + (-A) = (-A) + A = O$.

Let $A = \begin{bmatrix} 2 & -1 \\ 3 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 4 & 5 \\ -2 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 6 \\ 7 & -3 \end{bmatrix}$.

$$A + B = \begin{bmatrix} 6 & 4 \\ 1 & 1 \end{bmatrix}, \qquad B + A = \begin{bmatrix} 6 & 4 \\ 1 & 1 \end{bmatrix}$$

They are equal, as the commutative law says. For the associative law: $(A + B) + C = \begin{bmatrix} 6 & 10 \\ 8 & -2 \end{bmatrix}, \qquad A + (B + C) = \begin{bmatrix} 6 & 10 \\ 8 & -2 \end{bmatrix}$

Also $A + O = \begin{bmatrix} 2 & -1 \\ 3 & 0 \end{bmatrix} = A$ and $A + (-A) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.

Matrices are added entry by entry, so each rule for matrices comes from the same rule for the numbers inside them. For example, the entry in row $i$, column $j$ of $A + B$ is $a_{ij} + b_{ij}$. For $B + A$ it is $b_{ij} + a_{ij}$. These are equal because adding numbers can be done in either order. Since this is true for every position, $A + B = B + A$.

Let $A$ and $B$ be matrices of the same order and let $k$ and $l$ be scalars. Then:

• $k(A + B) = kA + kB$
• $(k + l)A = kA + lA$

Check. Let $A = \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 5 & 0 \\ 2 & -1 \end{bmatrix}$, $k = 3$ and $l = 2$. $k(A + B) = \begin{bmatrix} 18 & 12 \\ 0 & 6 \end{bmatrix}, \qquad kA + kB = \begin{bmatrix} 18 & 12 \\ 0 & 6 \end{bmatrix}$ $(k + l)A = \begin{bmatrix} 5 & 20 \\ -10 & 15 \end{bmatrix}, \qquad kA + lA = \begin{bmatrix} 5 & 20 \\ -10 & 15 \end{bmatrix}$ Each pair matches.

Matrix equations are solved just like ordinary equations. We may add or subtract the same matrix on both sides, and multiply both sides by the same number.

Example 1. Find the matrix $X$ if $2A + 3X = B$, where $A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 16 & -1 \\ 6 & 21 \end{bmatrix}$.

Solution. Subtract $2A$ from both sides: $3X = B - 2A$. Then multiply both sides by $\frac{1}{3}$: $X = \frac{1}{3}(B - 2A)$. Now $2A = \begin{bmatrix} 4 & 2 \\ 0 & 6 \end{bmatrix}$, so $B - 2A = \begin{bmatrix} 12 & -3 \\ 6 & 15 \end{bmatrix}$ and $X = \frac{1}{3}\begin{bmatrix} 12 & -3 \\ 6 & 15 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 2 & 5 \end{bmatrix}$

Example 2. Find $X$ and $Y$ if $X + Y = \begin{bmatrix} 8 & 6 \\ 4 & 4 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 4 & -2 \\ -6 & 2 \end{bmatrix}$.

Solution. Adding the two equations cancels $Y$: $2X = \begin{bmatrix} 12 & 4 \\ -2 & 6 \end{bmatrix}$, so $X = \begin{bmatrix} 6 & 2 \\ -1 & 3 \end{bmatrix}$. Subtracting the second equation from the first cancels $X$: $2Y = \begin{bmatrix} 4 & 8 \\ 10 & 2 \end{bmatrix}$, so $Y = \begin{bmatrix} 2 & 4 \\ 5 & 1 \end{bmatrix}$.

Example 3. Two stores sell three products. Their sales (in the same currency) in March and April are $M$ and $N$ (rows: stores; columns: products): $M = \begin{bmatrix} 1200 & 800 & 500 \\ 900 & 1100 & 700 \end{bmatrix}, \qquad N = \begin{bmatrix} 1500 & 600 & 800 \\ 1000 & 1300 & 400 \end{bmatrix}$ (a) Find the total sales for the two months. (b) Find the change from March to April. (c) The stores expect sales in the next period to be three times March's sales. Find that matrix.

Solution. (a) $M + N = \begin{bmatrix} 2700 & 1400 & 1300 \\ 1900 & 2400 & 1100 \end{bmatrix}$. (b) $N - M = \begin{bmatrix} 300 & -200 & 300 \\ 100 & 200 & -300 \end{bmatrix}$; a negative entry means sales fell. (c) $3M = \begin{bmatrix} 3600 & 2400 & 1500 \\ 2700 & 3300 & 2100 \end{bmatrix}$.

1. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $2A + 3A$ and $5A$. What do you notice?
2. Find $X$ if $X + \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}$.

Answers

1. $2A + 3A = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$ and $5A = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$. They are equal, as $(k + l)A = kA + lA$ says.
2. $X = \begin{bmatrix} 4 & 3 \\ 1 & 7 \end{bmatrix}$

Multiplying matrices is different from adding them. We do not multiply cell by cell. Instead, we combine a row of the first matrix with a column of the second. The easiest way to see why is to look at a bill.

A café sells coffee for 4, tea for 3 and juice for 5 (in the same currency). A customer orders 2 coffees, 1 tea and 3 juices. The bill is

$$2 \times 4 + 1 \times 3 + 3 \times 5 = 26$$

We can write the order as a row and the prices as a column. Multiplying the pairs and adding gives one number:

$$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 5 \end{bmatrix} = \begin{bmatrix} 26 \end{bmatrix}$$

A second customer orders 1 coffee, 4 teas and no juice. Their bill is $1 \cdot 4 + 4 \cdot 3 + 0 \cdot 5 = 16$. If we write each order as a row of one matrix $A$ (3 items), and keep the prices in a column, one multiplication gives both bills:

$$\underbrace{\begin{bmatrix} 2 & 1 & 3 \\ 1 & 4 & 0 \end{bmatrix}}_{2 \times 3} \underbrace{\begin{bmatrix} 4 \\ 3 \\ 5 \end{bmatrix}}_{3 \times 1} = \underbrace{\begin{bmatrix} 26 \\ 16 \end{bmatrix}}_{2 \times 1}$$

A second café charges 5 for coffee, 2 for tea and 4 for juice. Put the prices of the two cafés side by side as two columns of a matrix $B$. One multiplication now gives every bill at every café:

$$AB = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 4 & 0 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 3 & 2 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 26 & 24 \\ 16 & 13 \end{bmatrix}$$

Read the result like a table: row = which order, column = which café. For example, the entry 24 in row 1, column 2 is the bill for order 1 at the second café, $2 \cdot 5 + 1 \cdot 2 + 3 \cdot 4 = 24$.

Let $A$ be an $m \times n$ matrix and let $B$ be an $n \times p$ matrix. Then the product $AB$ is defined, and it is a matrix of order $m \times p$. Its entry in row $i$ and column $k$ is

$$c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k} + \dots + a_{in}b_{nk} = \sum_{j=1}^{n} a_{ij}b_{jk}$$

Example A. Find $AB$ if $A = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 0 & 5 \\ -3 & 2 \end{bmatrix}$.

Solution. $A$ is $2 \times 3$ and $B$ is $3 \times 2$. The middle numbers are both 3, so $AB$ is defined and its order is $2 \times 2$. Each entry is a row of $A$ times a column of $B$:

$$AB = \begin{bmatrix} 1 \cdot 2 + 2 \cdot 0 + 0 \cdot (-3) & 1 \cdot 1 + 2 \cdot 5 + 0 \cdot 2 \\ 3 \cdot 2 + (-1) \cdot 0 + 4 \cdot (-3) & 3 \cdot 1 + (-1) \cdot 5 + 4 \cdot 2 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ -6 & 6 \end{bmatrix}$$

Example B. In each case, say whether the product is defined. If it is, give its order.

Notice the last two rows. The same two lists of numbers give a single number in one order and a whole $3 \times 3$ table in the other order. This is the first sign that order matters in matrix multiplication.

Example C. A fitness app records the minutes a person spends on running, cycling and swimming on three days. Burning rates are 10, 8 and 9 calories per minute. Find the calories burned on each day.

Solution. Let $M$ be the minutes matrix ($3 \times 3$: days × activities) and $R$ the burning rates ($3 \times 1$: activities × calories per minute). Then $MR$ is $3 \times 1$, one total for each day:

$$MR = \begin{bmatrix} 30 & 0 & 20 \\ 0 & 45 & 30 \\ 20 & 20 & 0 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 9 \end{bmatrix} = \begin{bmatrix} 30 \cdot 10 + 0 \cdot 8 + 20 \cdot 9 \\ 0 \cdot 10 + 45 \cdot 8 + 30 \cdot 9 \\ 20 \cdot 10 + 20 \cdot 8 + 0 \cdot 9 \end{bmatrix} = \begin{bmatrix} 480 \\ 630 \\ 360 \end{bmatrix}$$

So the person burns 480, 630 and 360 calories on the three days.

1. For each pair, say whether the product is defined and give its order: (a) $3 \times 2$ and $2 \times 5$; (b) $4 \times 3$ and $4 \times 3$; (c) $1 \times 4$ and $4 \times 1$; (d) $5 \times 1$ and $1 \times 2$.
2. Find $AB$ if $A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 5 & -2 \end{bmatrix}$.
3. Find $CD$ if $C = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 3 & 4 \end{bmatrix}$ and $D = \begin{bmatrix} 1 & 2 \\ 0 & -3 \\ 5 & 1 \end{bmatrix}$. What is the order of $DC$?
4. A school canteen sells sandwiches, salads and soups. The quantities sold on two days are given by $Q = \begin{bmatrix} 12 & 8 & 5 \\ 9 & 10 & 7 \end{bmatrix}$ (rows: days; columns: sandwiches, salads, soups). The prices are 3, 4 and 2. Let $P = \begin{bmatrix} 3 \\ 4 \\ 2 \end{bmatrix}$ be the price column. Use matrix multiplication to find the income on each day.
5. Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Find $AB$ and $BA$. Are they equal?