Chapter 3 · Matrices

Exercise 3.1 · Q1: Order, number of elements and entries of a $3\times4$ matrix

Exercise 3.1Order of a matrix · Section 3.2

The question

In the matrix

A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \dfrac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix},

write:

(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements $a_{13}$ ⁠, $a_{21}$ ⁠, $a_{33}$ ⁠, $a_{24}$ ⁠, $a_{23}$ ⁠.

Exam answer

(i) $A$ has 3 rows and 4 columns, so its order is $3\times4$ .

(ii) Number of elements $= 3\times4$ $= 12$ .

(iii) $a_{13} = 19$ ⁠, $a_{21} = 35$ ⁠, $a_{33} = -5$ ⁠, $a_{24} = 12$ ⁠, $a_{23} = \dfrac52$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Every entry of a matrix has an address: the row it sits in and the column it sits in. The symbol $a_{ij}$ always means row $i$ first, then column $j$ . Counting rows and columns gives the order (rows × columns), and the number of elements is rows times columns. Everything in this question is reading that address system correctly.

Given, to find, and the plan

Given: The matrix $A$ above, with three horizontal lines of numbers and four vertical lines of numbers.
To find: (i) the order of $A$ ⁠; (ii) how many elements it has; (iii) the five entries $a_{13}$ ⁠, $a_{21}$ ⁠, $a_{33}$ ⁠, $a_{24}$ ⁠, $a_{23}$ ⁠.
Plan: (1) Count the rows and the columns. (2) Multiply them to get the number of elements, and confirm by counting. (3) For each $a_{ij}$ ⁠, go to row $i$ ⁠, then move along it to column $j$ ⁠.

Basics you need

1Rows and columns

The horizontal lines of entries are the rows (numbered 1, 2, 3, … from the top). The vertical lines are the columns (numbered 1, 2, 3, … from the left). Basics: rows, columns and the address

2Order of a matrix

A matrix with $m$ rows and $n$ columns has order $m\times n$ (read " $m$ by $n$ "). Rows are always named first. Lesson: what a matrix is

3Number of elements

An $m\times n$ matrix is a full rectangle: $m$ rows with $n$ entries in each, so it has $mn$ elements.

4The entry $a_{ij}$

$a_{ij}$ is the entry in row $i$ and column $j$ ⁠, also called the $(i, j)$ th element. The first subscript is the row, the second is the column. So $a_{24}$ and $a_{42}$ are different places. Basics: the address $a_{ij}$

Part (i): the order

Count the rows

Reading from top to bottom, $A$ has these horizontal lines:
$\begin{aligned} \text{row 1:}&\quad 2,\ 5,\ 19,\ -7 \\ \text{row 2:}&\quad 35,\ -2,\ \tfrac{5}{2},\ 12 \\ \text{row 3:}&\quad \sqrt{3},\ 1,\ -5,\ 17 \end{aligned}$
So $A$ has 3 rows.

The order is defined by the number of rows and the number of columns, so we must count both. Writing each row out separately makes it impossible to miscount.

Rows are horizontal and are numbered from the top. Basics: rows and columns
Count the columns

Each row contains 4 entries, so there are 4 vertical lines of entries:
$\begin{aligned} \text{column 1:}&\quad 2,\ 35,\ \sqrt{3} \\ \text{column 2:}&\quad 5,\ -2,\ 1 \\ \text{column 3:}&\quad 19,\ \tfrac{5}{2},\ -5 \\ \text{column 4:}&\quad -7,\ 12,\ 17 \end{aligned}$
So $A$ has 4 columns.

A matrix is a rectangular array (Definition 1), so every row has the same number of entries, and that common number is the number of columns. The fraction $\frac{5}{2}$ is one entry, not two: it sits in a single place.

Columns are vertical and are numbered from the left.
Write the order

Rows first, then columns: $A$ is of order $3\times 4$ .

The order $3\times4$ is not the same as $4\times3$ ⁠. A $4\times3$ matrix would have 4 rows and 3 columns, a different shape. The convention "rows first" is fixed by Section 3.2.1.

An $m\times n$ matrix has $m$ rows and $n$ columns. Lesson: order of a matrix

Part (ii): the number of elements

Multiply rows by columns

There are 3 rows and each row has 4 entries, so the number of elements is
$3 \times 4 = 12.$
Counting directly confirms it: rows 1, 2 and 3 each list 4 entries in Part (i), Step 1, and $4 + 4 + 4 = 12$ ⁠. So $A$ has 12 elements.

The rows are equal in length, so adding the row lengths is the same as multiplying: 3 rows of 4 is $3\times4$ ⁠. This is the general rule that an $m\times n$ matrix has $mn$ elements.

Repeated addition of equal amounts is multiplication: $4+4+4 = 3\times4$ ⁠.

Part (iii): the five entries

Read each address: row first, then column

For each $a_{ij}$ ⁠, go down to row $i$ ⁠, then along that row to its $j$ th entry. Use the rows listed in Part (i), Step 1.

Entry	Row $i$	Column $j$	Row $i$ is	Value
$a_{13}$	1	3	$2,\ 5,\ \mathbf{19},\ -7$	$19$
$a_{21}$	2	1	$\mathbf{35},\ -2,\ \tfrac52,\ 12$	$35$
$a_{33}$	3	3	$\sqrt3,\ 1,\ \mathbf{-5},\ 17$	$-5$
$a_{24}$	2	4	$35,\ -2,\ \tfrac52,\ \mathbf{12}$	$12$
$a_{23}$	2	3	$35,\ -2,\ \mathbf{\tfrac52},\ 12$	$\tfrac{5}{2}$

So $a_{13} = 19$ ⁠, $a_{21} = 35$ ⁠, $a_{33} = -5$ ⁠, $a_{24} = 12$ ⁠, $a_{23} = \frac{5}{2}$ .

A

as a grid of addresses. The shaded cells are the five entries asked for. Notice that

a_{23}

and

a_{24}

are neighbours in row 2, while

a_{13}

⁠,

a_{23}

⁠,

a_{33}

run down column 3.

The subscripts are an address, and the order of the two numbers matters. Reading

a_{21}

as "column 2, row 1" would give

5

instead of

35

⁠. Swapping the subscripts changes four of these five answers:

a_{31} = \sqrt3

(not 19),

a_{12} = 5

(not 35),

a_{32} = 1

(not

\frac52

⁠), and

a_{42}

does not even exist, because

A

has no row 4. Only

a_{33}

survives the swap, because its two subscripts are the same.

a_{ij}

⁠: first subscript = row, second subscript = column. Basics: the address

a_{ij}

Result

(i) The order of $A$ is $3\times4$ ⁠. (ii) $A$ has $12$ elements. (iii) $a_{13} = 19$ ⁠, $a_{21} = 35$ ⁠, $a_{33} = -5$ ⁠, $a_{24} = 12$ ⁠, $a_{23} = \frac{5}{2}$ ⁠. (Textbook answer: (i) $3\times4$ ⁠, (ii) 12, (iii) $19, 35, -5, 12, \frac52$ ⁠.)

Check it with numbers

Every entry of $A$ has exactly one address $(i, j)$ with $i$ in $\{1,2,3\}$ and $j$ in $\{1,2,3,4\}$ ⁠. There are 3 choices for $i$ and 4 for $j$ ⁠, so $3\times 4 = 12$ addresses, matching Part (ii) ✓. Column 3 read downwards is $a_{13}, a_{23}, a_{33}$ ⁠, which is $19, \frac52, -5$ ⁠: the same three numbers found in the table ✓.

Common mistakes

Writing the order as $4\times3$ ⁠. The order is rows × columns; $A$ has 3 rows.
Swapping the subscripts. $a_{21}$ is row 2, column 1, which is $35$ ⁠, not $5$ ⁠.
Counting $\frac52$ or $\sqrt3$ as more than one element. Each is a single number in a single place.
Writing $a_{24} = -7$ ⁠. $-7$ is in column 4 of row 1, so it is $a_{14}$ ⁠.

What to take away

Order = (number of rows) × (number of columns), rows first.
An $m\times n$ matrix has $mn$ elements.
$a_{ij}$ is found by going to row $i$ ⁠, then along to column $j$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
A matrix is an ordered rectangular array; rows are horizontal, columns vertical	Part (i), Steps 1, 2	Definition 1, Section 3.2 (Lesson: what a matrix is)
An $m\times n$ matrix has $m$ rows and $n$ columns	Part (i), Step 3	Section 3.2.1, Order of a matrix
An $m\times n$ matrix has $mn$ elements	Part (ii), Step 1	Section 3.2.1
$a_{ij}$ is the element in the $i$ th row and $j$ th column	Part (iii), Step 1	Section 3.2.1
All entries and counts	Throughout	Recomputed in the site's verification script

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Exercise 3.1 · Q1: Order, number of elements and entries of a 3×43\times43×4 matrix

Basics you need

Part (i): the order

Count the rows

Count the columns

Write the order

Part (ii): the number of elements

Multiply rows by columns

Part (iii): the five entries

Read each address: row first, then column

Every fact used, and where it comes from

Exercise 3.1 · Q1: Order, number of elements and entries of a $3\times4$ matrix