📐 Matrices — Revision Notes

A matrix is information laid out in a rectangle of brackets: the rows mean one thing, the columns another. Adding and scaling go element by element; multiplying goes row by column. The sentence that says what the answer's entries mean carries marks of its own. Syllabus outcomes N9 9.1, 9.2, 9.3 and 9.4.

What this chapter is ✕ Not in G2

Formula sheet: no matrices Exact answers; money to 2 d.p. Show the unsimplified matrix

A matrix (plural matrices) is a rectangular array of numbers written inside a pair of brackets. The numbers are its elements. A matrix with m rows and n columns is said to have order m × n — rows first, always.

The question	Where it is answered	The tool
What is its order, and what does that entry mean?	6.1 What a Matrix Is	`m` rows × `n` columns, and the row and column headings
How do I add, subtract or scale them?	6.2 Adding, Subtracting & Scaling	element by element — but only if the orders are identical
How do I multiply two matrices, and when can I?	6.3 Multiplying Two Matrices	row × column, after the inner dimensions have been checked
The question is about money, totals or points. Which product?	6.4 Matrices as Data Models	quantities × prices — then say what each entry represents

Where this chapter stops. N9 asks four things: display information as a matrix, interpret it, multiply by a scalar, and add, subtract or multiply matrices. Inverses, determinants and the transpose belong to Additional Mathematics; the boundary is drawn in 6.3.

More detail

K310 says "Relevant mathematical formulae will be provided", and the printed MATHEMATICAL FORMULAE page does hold compound interest, mensuration, ½ab sin C, arc length and sector area in radians, the sine and cosine rules, the mean and the standard deviation — and not one word about matrices. The order convention, the conformability test and the row-times-column rule are carried in your head; there are only three of them. Two rules from the front of the paper, quoted: "Omission of essential working will result in loss of marks." and "Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question." Matrix arithmetic is exact — every entry here is a whole number, a decimal or a fraction that ends — so almost nothing in this chapter is rounded at all. The one exception is money, which is written to 2 decimal places. What "essential working" means here is the unsimplified matrix: the one whose entries are still written as 6 × 4 + 3 × 5 rather than 39. There is also no algebra of the identity matrix in this syllabus. If you meet

M^{- 1}

or det M in a book, that is Additional Mathematics material rather than anything N9 asks for.

Everything in this tab is examined in G3 Mathematics only. ✕ Not in G2

Order, rows, columns and elements

A matrix is named with a capital letter in bold in print — A, M, P — which you write with an ordinary capital by hand. To point at one particular element, use a small letter with two subscripts: row number first, column number second.

In a matrix A, the element

a_{i j}

is the element in row i and column j. So

a_{23}

is the element in row 2, column 3 — and it is in general not the same element as

a_{32}

Fig. 6.1

The rows run across and the columns run down — always, in every matrix, whatever the word "row" means when a teacher lines a class up. This matrix has 2 rows and 3 columns, so its order is 2 × 3. It has 2 × 3 = 6 elements.

Rows first. Order is written m × n: m rows, n columns. A 2 × 3 matrix and a 3 × 2 matrix are different objects, and 6.3 turns on reading the pair in that order.

For the canteen matrix above — call it M — $m_{13}$ = 32: row 1 is Stall 1, column 3 is Milo, so 32 cups of Milo were sold at Stall 1.

Walkthrough 1 — stating the order Basic

Write down the order of each of the following matrices. (a)

(\begin{matrix} 5 & −2 & 0 \\ 7 & 1 & 4 \end{matrix})

(b)

(\begin{matrix} 8 \\ −3 \\ 6 \\ 1 \end{matrix})

(c)

(\begin{matrix} −6 & 9 \end{matrix})

(d)

(\begin{matrix} 3 & 0 \\ −5 & 2 \end{matrix})

(a) 2 rows, 3 columns → order 2 × 3

Count the horizontal lines of numbers first: there are two. Then count how many numbers are in one of those lines: three. Rows first, so 2 × 3. The minus sign on the −2 is part of that element, not an extra one.
(b) 4 rows, 1 column → order 4 × 1

Four numbers stacked vertically make four rows of one element each, not one row of four. A matrix with a single column like this is called a column matrix.
(c) 1 row, 2 columns → order 1 × 2

The mirror image of (b): one horizontal line, two elements in it — a row matrix. Elements are separated by a space, not a comma, which in some countries' notation would read as a decimal point.
(d) 2 rows, 2 columns → order 2 × 2
So the orders are 2 × 3, 4 × 1, 1 × 2 and 2 × 2.

A matrix with as many rows as columns is a square matrix, and (d) is the only square one here. Write the answers as orders, with the multiplication sign — "2 by 2", not "4 elements".

Where this comes from

The figure starts from two drink stalls in a canteen counting what they sold one morning. Written as a table it is six numbers with six labels; written as a matrix it is six numbers and one shape, and the shape is what every later rule in this chapter is about.

Reading the elements of M in order means reading the whole of row 1 from the left, then the whole of row 2: 24, 17, 32, 41, 9, 28.

The four names worth knowing

None of these is a new idea — each is just a shape of matrix that comes up often enough to have earned a name, and questions use the names without explaining them.

Name	Shape	Example
Row matrix	one row, any number of columns (1 × `n`)	$(\begin{matrix} 4 & −1 & 7 \end{matrix})$
Column matrix	one column, any number of rows (`m` × 1)	$(\begin{matrix} 4 \\ −1 \\ 7 \end{matrix})$
Square matrix	as many rows as columns (`n` × `n`)	$(\begin{matrix} 4 & −1 \\ 7 & 0 \end{matrix})$
Zero matrix (or null matrix), written 0	every element is 0 — of any order	$(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$

A 1 × 1 matrix is not a number. The matrix

(\begin{matrix} 7 \end{matrix})

and the number 7 are different objects, and so are the zero matrix 0 and the number 0. A 1 × 1 answer keeps its brackets.

More detail

It matters in exactly one place, and there is a mark in it: a product such as $(\begin{matrix} 2 & 5 \end{matrix})$ $(\begin{matrix} 3 \\ 1 \end{matrix})$ comes out as the 1 × 1 matrix $(\begin{matrix} 11 \end{matrix})$ , written with its brackets. When the same question then asks "what is the total cost?", that answer is the plain number 11 with a dollar sign.

Reading a matrix — what the entries mean

Outcome 9.2 is "interpreting the data in a given matrix": you say, in a sentence, what a number stands for. A matrix on its own is ambiguous, so label it — write the row and column headings outside the brackets, the way the figure above does.

Three questions answer almost every "what does this represent?" part:
1. Which row is it in, and what does that row stand for?
2. Which column is it in, and what does that column stand for?
3. What are the units — cups, dollars, points, people?

Walkthrough 2 — from a table to a matrix, and back to a sentence Basic

A school library records how many items were borrowed on two days. On Monday: 46 fiction books, 23 non-fiction books and 19 magazines. On Tuesday: 38 fiction books, 31 non-fiction books and 12 magazines. (i) Represent the data by a matrix L, and state its order. (ii) Write down the value of

l_{21}

and say what it represents. (iii) Find the sum of the elements in the first row, and say what this sum represents.

$L = (\begin{matrix} 46 & 23 & 19 \\ 38 & 31 & 12 \end{matrix})$ rows: Monday, Tuesday | columns: fiction, non-fiction, magazines

Two days and three kinds of item, so one day per row and one kind per column. The headings are written beside the matrix even though the question does not ask for them: they are what makes each entry readable.
The order of L is 2 × 3.

Two rows (the two days), three columns (the three kinds of item) — rows first. Transposing the table, items down the side and days across the top, would have given a 3 × 2 matrix — just as correct, provided the headings said so.
$l_{21}$ = 38: the number of fiction books borrowed on Tuesday.

Row 2 is Tuesday, column 1 is fiction. Both halves of the sentence are needed — "38 fiction books" without the day, or "38 borrowed on Tuesday" without the kind, only answers half the question.
46 + 23 + 19 = 88: the total number of items borrowed on Monday.

Row 1 is Monday, and running along a row crosses all three columns, so the sum covers every kind of item on that one day. Name the day and say "items", not "books" — magazines are in the total too.

More detail

24 is only "24 cups of Kopi at Stall 1" once somebody has said which way round the rows and the columns run; with the headings written down, the interpretation writes itself.

The same three questions handle sums of entries. The sum of a row is a total over everything the columns stand for (all the drinks at one stall); the sum of a column is a total over everything the rows stand for (Kopi at every stall). Which of those two a question wants is decided by the labels, not by the arithmetic — and 6.4 shows how to make matrix multiplication produce both of them for you.

When are two matrices equal?

Two matrices A and B are equal if and only if (a) they have the same order, and (b) their corresponding elements are equal — that is, the element in each position of A equals the element in the same position of B.

Equality is what turns a matrix into equations. An equation between two matrices of order m × n is really mn ordinary equations stacked up — one per position. The same idea answers the "find the unknowns" questions in 6.2 and 6.3.

Walkthrough 3 — equal matrices give you equations Intermediate

Given that

A = (\begin{matrix} 3 a & b \\ c & 12 \end{matrix})

and

B = (\begin{matrix} 27 & a + 5 \\ d - 7 & d \end{matrix})

and that A = B, find the values of a, b, c and d.

Both matrices are 2 × 2, so equating corresponding elements:
3a = 27 b = a + 5 c = d − 7 12 = d

Same order, so the equality is legal; four positions, so four equations. Writing all four out first is the "essential working" here — pick them off position by position, going along row 1 and then row 2, so that none is missed.
From 3a = 27: a = 9

Only one of the four equations has a single unknown in it, so it is the one to start with. Everything else follows from a or from d.
b = a + 5 = 9 + 5 = 14

Substitute the value of a just found. The unknown a appears on both sides of this equation — in A as 3a, in B as a + 5 — so the equations are solved in order.
From 12 = d: d = 12

The bottom-right position gives d immediately. A number on the left and a letter on the right is still an equation; there is nothing to do but read it.
c = d − 7 = 12 − 7 = 5
a = 9, b = 14, c = 5, d = 12

List all four values, in the order the question named them. A quick check costs five seconds: both matrices become $(\begin{matrix} 27 & 14 \\ 5 & 12 \end{matrix})$ , so they really are equal.

More detail

Both halves of the definition matter. $(\begin{matrix} 3 & 0 \end{matrix})$ and $(\begin{matrix} 3 \\ 0 \end{matrix})$ contain the same two numbers, but one is 1 × 2 and the other is 2 × 1, so they are not equal. Neither are $(\begin{matrix} 1 & 5 \\ 2 & 6 \end{matrix})$ and $(\begin{matrix} 1 & 2 \\ 5 & 6 \end{matrix})$ : same four numbers, same order, different positions.

Check yourself

Write down the order of $(\begin{matrix} −4 & 7 \\ 0 & 3 \\ 5 & −1 \end{matrix})$ .
Answer
Three horizontal lines of numbers, two numbers in each, so the order is 3 × 2 — rows first.
A matrix has 12 elements and 3 rows. How many columns does it have, and what is its order?
Answer
12 ÷ 3 = 4 columns, so the order is 3 × 4. The number of elements is always (rows) × (columns), which is where the × in the order comes from.
Given $G = (\begin{matrix} 6 & −2 & 9 \\ 0 & 5 & −7 \end{matrix})$ , write down $g_{23}$ and $g_{12}$ .
Answer
$g_{23}$ is row 2, column 3 = −7. $g_{12}$ is row 1, column 2 = −2. Row number first, both times.
Are $(\begin{matrix} 3 & 0 \end{matrix})$ and $(\begin{matrix} 3 \\ 0 \end{matrix})$ equal? Give a reason.
Answer
No. The first is a 1 × 2 row matrix and the second is a 2 × 1 column matrix. Equality requires the same order before the elements are even looked at.
Find x and y given that $(\begin{matrix} 2 x & 7 \\ 5 & y + 1 \end{matrix})$ = $(\begin{matrix} 14 & 7 \\ 5 & −3 \end{matrix})$ .
Answer
2x = 14 gives x = 7; y + 1 = −3 gives y = −4. The other two positions (7 = 7 and 5 = 5) carry no information — they are still worth a glance, because if one of them had been false the matrices could not have been equal at all.
Is the 1 × 1 matrix $(\begin{matrix} 9 \end{matrix})$ the same thing as the number 9?
Answer
No. One is a matrix — an array with an order — and the other is a number. They behave differently: a matrix can only be added to another matrix of the same order, whereas 9 can be added to anything. In practice this only bites when a product turns out to be 1 × 1: write the brackets in the matrix line, and drop them only when you state the answer in context.

Everything in this tab is examined in G3 Mathematics only. ✕ Not in G2

Before you multiply: does the product even exist?

Each entry of a product is built from a whole row of the first matrix and a whole column of the second, so the row and the column being paired have to be the same length. The product AB therefore exists only if the number of columns of A equals the number of rows of B.

The conformability test. Write the two orders side by side: the inner pair must match, and the outer pair gives the order of the answer.

Fig. 6.2

Write the two orders down before writing anything else. If the inner pair does not match, stop — the product does not exist, and saying so is the answer. If it does match, the outer pair has already told you how many entries you are about to compute, which is the best check there is against losing one.

Two habits that follow straight from the test:
1. AB existing does not mean BA exists. A 2 × 3 times a 3 × 4 is fine; the same two the other way round is 3 × 4 times 2 × 3, whose inner pair is 4 and 2. No product.
2. When both do exist they may have different orders — see Walkthrough 11, where the same two matrices give a 1 × 1 answer one way round and a 3 × 3 answer the other.

Walkthrough 9 — which products exist, and of what order Basic

A = (\begin{matrix} 2 & −5 & 1 \\ 0 & 3 & 4 \end{matrix})

B = (\begin{matrix} 7 & −1 \\ 0 & 4 \\ 5 & 2 \end{matrix})

C = (\begin{matrix} 3 & 6 \\ −2 & 1 \end{matrix})

D = (\begin{matrix} 8 & −3 & 5 \end{matrix})

. State whether each of the following products exists, and if it does, write down its order. (a) AB (b) BA (c) AC (d) DB

A is 2 × 3, B is 3 × 2, C is 2 × 2, D is 1 × 3.

Write every order down before looking at a single part. Four orders cost ten seconds and turn the rest of the question into pattern-matching; hunting through the matrices again for each part is how the wrong dimension gets used.
(a) (2 × 3)(3 × 2): the inner pair matches, so AB exists and its order is 2 × 2.

A has 3 columns and B has 3 rows, so each row of A is exactly long enough to pair with each column of B. The outer pair, 2 and 2, gives the order.
(b) (3 × 2)(2 × 3): the inner pair matches, so BA exists and its order is 3 × 3.

Both products exist here, but they are not the same size: AB is 2 × 2 and BA is 3 × 3. Matrices of different orders cannot be equal, so the two products need not agree.
(c) (2 × 3)(2 × 2): 3 ≠ 2, so AC does not exist.

A row of A has three numbers and a column of C only two, so there is nothing to pair the third with. Say why: the number of columns of A is not equal to the number of rows of C.
(d) (1 × 3)(3 × 2): the inner pair matches, so DB exists and its order is 1 × 2.
So AB is 2 × 2, BA is 3 × 3, AC does not exist, and DB is 1 × 2.

A row matrix times a matrix gives a row matrix — the outer pair is 1 and 2. Collect the four answers together at the end so that nothing is left as a loose comment in the margin.

Where this comes from

Addition and subtraction were the obvious rules — element to matching element. Matrix multiplication is not the obvious rule: you do not multiply matching elements. Because each entry of the answer instead consumes a whole row and a whole column, the shapes have to agree before any arithmetic can start — which is why the order check comes before the multiplying, and not after it.

How the multiplication actually works

Once the orders have been checked, every entry of the answer is built the same way: the entry of AB in row i, column j comes from row i of A and column j of B, multiplied together term by term and then added. Fill the first row of the answer completely before starting the second.

Here it is on a 2 × 3 times a 3 × 2, which by the conformability test must give a 2 × 2 answer. The shaded entry is being built from the shaded row and the shaded column.

Fig. 6.3

Row 2 of P is 0, 4, −2 and column 1 of Q is 1, −3, 4. Pair them in order — first with first, second with second, third with third — multiply, and add: 0 − 12 − 8 = −20. Because it came from row 2 and column 1, it is the entry in row 2, column 1 of the answer.

The other three entries are built the same way, and every one of them is a sum of three products, because the rows of P and the columns of Q are three numbers long:

Entry	Row × column	Value
row 1, column 1	(2)(1) + (−1)(−3) + (3)(4) = 2 + 3 + 12	17
row 1, column 2	(2)(5) + (−1)(2) + (3)(0) = 10 − 2 + 0	8
row 2, column 1	(0)(1) + (4)(−3) + (−2)(4) = 0 − 12 − 8	−20
row 2, column 2	(0)(5) + (4)(2) + (−2)(0) = 0 + 8 + 0	8

Collecting the four entries:

(\begin{matrix} 2 & −1 & 3 \\ 0 & 4 & −2 \end{matrix}) (\begin{matrix} 1 & 5 \\ −3 & 2 \\ 4 & 0 \end{matrix}) = (\begin{matrix} 17 & 8 \\ −20 & 8 \end{matrix})

Two habits that stop nearly every lost mark here.
1. Write the unsimplified matrix first — the one whose entries are still "(2)(1) + (−1)(−3) + (3)(4)". It is the essential working, it is where the method mark lives, and it separates the pairing from the arithmetic so that a slip in one does not hide a slip in the other.
2. Keep every substituted number in its own brackets, so that (−1)(−3) cannot be read as −1 − 3.

Walkthrough 10 — a first product Basic

Given

A = (\begin{matrix} 3 & −2 \\ 1 & 4 \end{matrix})

and

B = (\begin{matrix} 5 & 0 \\ −1 & 6 \end{matrix})

, evaluate AB.

(2 × 2)(2 × 2): the inner pair matches, so AB exists and is 2 × 2 — four entries to find.

The check comes first, every time, even when both matrices are obviously square. Knowing the answer is 2 × 2 also tells you how many entries to write, so a missing one is noticed immediately.
Row 1 of A is (3 −2). Pairing it with column 1 and then column 2 of B:
(3)(5) + (−2)(−1) and (3)(0) + (−2)(6)

Column 1 of B is (5, −1) read downwards and column 2 is (0, 6). Fill the whole first row of the answer before touching row 2 — working in that order is what stops entries landing in the wrong place.
Row 2 of A is (1 4), giving
(1)(5) + (4)(−1) and (1)(0) + (4)(6)

Same two columns of B, now paired with the second row of A. Notice that B's columns get used again — every column of the second matrix is used once for every row of the first.
AB = $(\begin{matrix} 15 + 2 & 0 - 12 \\ 5 - 4 & 0 + 24 \end{matrix}) = (\begin{matrix} 17 & −12 \\ 1 & 24 \end{matrix})$

(−2)(−1) = +2, so the top-left entry is 15 + 2 = 17 — the double negative is the one to watch. The answer is 2 × 2, as predicted, and it is written inside brackets.

Walkthrough 11 — the same two matrices, two very different answers Intermediate

Given

R = (\begin{matrix} 4 & −2 & 5 \end{matrix})

and

S = (\begin{matrix} 3 \\ 7 \\ −1 \end{matrix})

, evaluate (a) RS and (b) SR.

R is 1 × 3 and S is 3 × 1.
RS: (1 × 3)(3 × 1) → 1 × 1. SR: (3 × 1)(1 × 3) → 3 × 3.

Both inner pairs match, so both products exist — but the outer pairs are 1, 1 and 3, 3. One answer is a single bracketed number, the other a block of nine, so the number of entries to write is known before any arithmetic.
(a) RS = $(\begin{matrix} (4) (3) + (−2) (7) + (5) (−1) \end{matrix}) = (\begin{matrix} −7 \end{matrix})$

There is one row and one column, so there is one entry: 12 − 14 − 5 = −7. Keep the brackets — the answer is the 1 × 1 matrix (−7), not the number −7, until a question in context asks for a total.
(b) SR = $(\begin{matrix} (3) (4) & (3) (−2) & (3) (5) \\ (7) (4) & (7) (−2) & (7) (5) \\ (−1) (4) & (−1) (−2) & (−1) (5) \end{matrix})$

Now each row of S is a single number and each column of R is a single number, so every entry is a sum of just one product — no adding at all. Nine entries, because the answer is 3 × 3.
SR = $(\begin{matrix} 12 & −6 & 15 \\ 28 & −14 & 35 \\ −4 & 2 & −5 \end{matrix})$
So RS = (−7), a 1 × 1 matrix, while SR is the 3 × 3 matrix above.

Two matrices, two orders of multiplication, and the answers are not even the same shape. That is why a question never says simply "multiply R and S" — it says which one is written first.

`AB` is not `BA`

For ordinary numbers, 6 × 7 and 7 × 6 are the same. Matrix multiplication is not commutative: in general $A B \neq B A$ , even when both products exist and both are square of the same order.

A question can hinge entirely on this: "explain why…", "show that…", or "find the value of k for which AB = BA" — that last one has an answer only because equality is the exception.

Walkthrough 12 — a counter-example where both products exist Intermediate

Given

A = (\begin{matrix} 1 & 2 \\ 3 & 0 \end{matrix})

and

B = (\begin{matrix} 4 & −1 \\ 2 & 5 \end{matrix})

, show that AB ≠ BA.

Both are 2 × 2, so both AB and BA exist and both are 2 × 2.

Say this first. If the two products had different orders there would be nothing to show — they could not be equal for a trivial reason. Here they are the same shape, so the difference is found entry by entry.
AB = $(\begin{matrix} 4 + 4 & −1 + 10 \\ 12 + 0 & −3 + 0 \end{matrix}) = (\begin{matrix} 8 & 9 \\ 12 & −3 \end{matrix})$

Rows of A against columns of B: (1)(4) + (2)(2) = 8, (1)(−1) + (2)(5) = 9, (3)(4) + (0)(2) = 12, (3)(−1) + (0)(5) = −3. The zero in A kills one product in each bottom entry, not the entry itself.
BA = $(\begin{matrix} 4 - 3 & 8 + 0 \\ 2 + 15 & 4 + 0 \end{matrix}) = (\begin{matrix} 1 & 8 \\ 17 & 4 \end{matrix})$

Now the rows come from B and the columns from A: (4)(1) + (−1)(3) = 1, (4)(2) + (−1)(0) = 8, (2)(1) + (5)(3) = 17, (2)(2) + (5)(0) = 4. The same eight numbers, paired differently.
Since $(\begin{matrix} 8 & 9 \\ 12 & −3 \end{matrix})$ ≠ $(\begin{matrix} 1 & 8 \\ 17 & 4 \end{matrix})$ , AB ≠ BA.

A "show that" ends with a written conclusion, not two matrices left side by side. One position differing is enough — here all four do, and the top-left entries, 8 and 1, settle it.

Walkthrough 13 — making AB = BA happen Advanced

A = (\begin{matrix} 3 & 0 \\ 2 & 1 \end{matrix})

and

B = (\begin{matrix} 1 & 0 \\ k & 4 \end{matrix})

. Find the value of k for which AB = BA.

AB = $(\begin{matrix} 3 + 0 & 0 + 0 \\ 2 + k & 0 + 4 \end{matrix}) = (\begin{matrix} 3 & 0 \\ 2 + k & 4 \end{matrix})$

Rows of A against columns of B, with k carried along like any other number. Only the bottom-left entry contains k: (2)(1) + (1)(k) = 2 + k, because k only ever meets the 0 in the top row.
BA = $(\begin{matrix} 3 + 0 & 0 + 0 \\ 3 k + 8 & 0 + 4 \end{matrix}) = (\begin{matrix} 3 & 0 \\ 3 k + 8 & 4 \end{matrix})$

Now the rows come from B: the bottom-left entry is (k)(3) + (4)(2) = 3k + 8. The same entry carries k, now with a coefficient of 3 and an extra 8 — the difference between the two orders.
Equating corresponding elements:
2 + k = 3k + 8

Three of the four positions already agree (3 = 3, 0 = 0, 4 = 4), so they place no condition on k. The whole question comes down to the one position that differs.
2 − 8 = 3k − k
−6 = 2k, so k = −3

With k = −3 the bottom-left entries become 2 + (−3) = −1 and 3(−3) + 8 = −1, which agree, so both products equal $(\begin{matrix} 3 & 0 \\ −1 & 4 \end{matrix})$ . For every other value of k these two matrices do not commute.

Where matrix algebra stops. You now know three operations — addition, subtraction and multiplication — plus the scalar multiple. There is no division: a matrix question in this course that looks as though it needs division needs the right product, the right way round.

More detail

In ordinary algebra you would undo a multiplication by dividing; with matrices the corresponding object is the inverse matrix $M^{- 1}$ , and finding one needs the determinant — neither of which is in N9. Nor is the transpose, and nor is any algebra built on the identity matrix $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ .

Those symbols are printed in K310's Mathematical Notation appendix, but that list opens by saying it is "primarily directed towards A-Level" and applies to other levels only "where relevant" — the same list also carries complex numbers. The content table is the one that decides, and its N9 row has exactly four bullets. Three operations and a scalar multiple are the whole toolkit.

Check yourself

A is a 4 × 2 matrix and B is a 2 × 5 matrix. State the order of AB, and say what happens with BA.
Answer
(4 × 2)(2 × 5): the inner pair matches, so AB exists and is 4 × 5. For BA: (2 × 5)(4 × 2) — 5 ≠ 4, so BA does not exist.
Evaluate $(\begin{matrix} 2 & 1 \\ 0 & 3 \end{matrix}) (\begin{matrix} 4 & −1 \\ 5 & 2 \end{matrix})$ .
Answer
(2)(4) + (1)(5) = 13; (2)(−1) + (1)(2) = 0; (0)(4) + (3)(5) = 15; (0)(−1) + (3)(2) = 6, giving $(\begin{matrix} 13 & 0 \\ 15 & 6 \end{matrix})$ .
Evaluate $(\begin{matrix} 3 & −1 & 2 \end{matrix}) (\begin{matrix} 5 \\ 0 \\ −4 \end{matrix})$ , and state the order of your answer.
Answer
(1 × 3)(3 × 1) gives a 1 × 1 matrix. (3)(5) + (−1)(0) + (2)(−4) = 15 + 0 − 8 = 7, so the answer is $(\begin{matrix} 7 \end{matrix})$ — with its brackets.
Evaluate $(\begin{matrix} 5 \\ −2 \end{matrix}) (\begin{matrix} 3 & 1 \end{matrix})$ .
Answer
(2 × 1)(1 × 2) gives a 2 × 2 matrix, each entry a single product: $(\begin{matrix} 15 & 5 \\ −6 & −2 \end{matrix})$ . Compare it with the previous question: a column times a row grows, a row times a column collapses.
Given $A = (\begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix})$ and $B = (\begin{matrix} 0 & 2 \\ 4 & 1 \end{matrix})$ , show that AB ≠ BA.
Answer
AB = $(\begin{matrix} 12 & 5 \\ 0 & 4 \end{matrix})$ and BA = $(\begin{matrix} 4 & 0 \\ 6 & 12 \end{matrix})$ . These are not equal (the top-left entries are 12 and 4), so AB ≠ BA. Finish with that sentence.
Given that $(\begin{matrix} 2 & 3 \\ 1 & h \end{matrix}) (\begin{matrix} 4 \\ 2 \end{matrix}) = (\begin{matrix} 14 \\ 8 \end{matrix})$ , find h.
Answer
The product is (2 × 2)(2 × 1), so it is 2 × 1 as required. Top: (2)(4) + (3)(2) = 14, which checks out and gives no information. Bottom: (1)(4) + (h)(2) = 8, i.e. 4 + 2h = 8, so h = 2.

The whole toolkit, on one page ✕ Not in G2

Four operations, one condition each, and one sentence at the end. None of it is on the formula sheet.

1. Order and elements. A matrix with m rows and n columns has order m × n — rows first — and mn elements. The element

a_{i j}

sits in row i, column j.

Equality. Two matrices are equal only if they have the same order and every pair of corresponding elements is equal — which turns one matrix equation into mn ordinary ones.

2. Addition and subtraction. Only for identical orders, and then element by corresponding element; the answer keeps that order. Different orders → the sum or difference does not exist, and saying so with the reason is the answer.

Subtraction runs element by corresponding element in the same way:

(\begin{matrix} a & b \\ c & d \end{matrix}) - (\begin{matrix} e & f \\ g & h \end{matrix}) = (\begin{matrix} a - e & b - f \\ c - g & d - h \end{matrix})

3. Scalar multiple. kA multiplies every element of A by k — zeros and negatives included. There is no condition to check, and the order is unchanged. Write the scalar in front: 2A, not A2.

4. Multiplication — the check. (m × n)(n × p): the inner pair must match, and the outer pair gives the order m × p.

4. Multiplication — the rule. The entry in row i, column j of AB comes from row i of A and column j of B, multiplied term by term and added. Fill each row of the answer before starting the next.

5. $A B \neq B A$ . Matrix multiplication is not commutative: the two products may not both exist; if they both exist they may have different orders; and even when both are square of the same order they are usually different.

A question that says "find k such that AB = BA" has an answer because equality is the exception, not the rule.

6. What the answer's entries mean. The rows of AB carry A's row label, the columns carry B's column label, and the label that A's columns shared with B's rows has been summed away.

That is the whole of "interpret the data": name the row, name the column, give the units.

Which operation, and what it needs ✕ Not in G2

You are asked to…	It needs	The answer's order	Where the marks go
State the order	nothing	—	rows × columns, in that order — "2 × 3", not "6 elements"
Say what an element represents	the row and column headings	—	a sentence naming the row, the column and the units
Use `A` = `B` to find unknowns	identical orders	—	"equating corresponding elements", then one equation per position
`A` + `B` or `A` − `B`	identical orders	the same order as `A`	the unsimplified matrix; brackets round every negative
`kA`	nothing — any order works	the same order as `A`	every element multiplied, including 0 and the negatives
`hA` + `kB`	`A` and `B` of identical order	that same order	both scalars multiplied out before the addition line
`AB`	columns of `A` = rows of `B`	(rows of `A`) × (columns of `B`)	the order check first, then the unsimplified matrix
Show `AB` ≠ `BA`	both products to exist	may differ between the two	both products in full, then a written conclusion
"Not possible" questions	the order check	—	the reason: which dimension failed to match which
Total cost / total takings	quantities × prices, prices as a column	(number of shops) × 1	choosing the order; then the money to 2 d.p. in a sentence
Row totals / column totals	a column of 1s after, or a row of 1s before	`m` × 1, or 1 × `n`	the right number of 1s, and what the totals are totals of

The habits that earn the marks

1. Write the order of every matrix in the question before doing anything with it — on the paper, next to the matrix.
2. For any product, do the inner/outer check first and write down the order of the answer. It tells you how many entries to expect.
3. Show the unsimplified matrix — the one still full of "(3)(5) + (−2)(−1)". That line is the method mark.
4. Keep every substituted negative inside its own brackets.
5. Fill the answer row by row, left to right, so nothing lands in the wrong place.
6. Never swap the order of a product to make it easier — the answer changes.
7. If something cannot be done, write the sentence with the reason in it. A blank scores nothing and an invented matrix scores less.
8. Finish an applied question by saying what the entries represent, with units — and turn a 1 × 1 answer into a plain amount of money when the question asked for a total.

Syllabus at a glance

The four G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus — and the reason the G2 panel beside it is empty.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
9.1	display of information in the form of a matrix of any order ✕ Not in G2	6.1 Order and elements, 6.1 Kinds of matrix
9.2	interpreting the data in a given matrix ✕ Not in G2	6.1 Reading a matrix, 6.4 Data models, 6.4 Totals
9.3	product of a scalar quantity and a matrix ✕ Not in G2	6.2 Scalar multiples, 6.2 Matrix equations
9.4	problems involving addition, subtraction and multiplication of matrices ✕ Not in G2	6.2 Addition & subtraction, 6.3 Conformability, 6.3 The rule, 6.3 `AB` ≠ `BA`, 6.4 Two-stage products

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus. The K310 content table words the last one slightly differently — "problems involving the calculation of the sum and product (where appropriate) of two matrices" — and the parenthesis is worth reading twice: it is the conformability caveat, and it is the syllabus itself telling you that the order check is part of the outcome rather than a preliminary to it. Three things this chapter uses are not separately listed under N9. The equality of two matrices (6.1) is the tool that turns 9.4's "problems" into equations; the $a_{i j}$ element notation (6.1) is how 9.2's "interpreting the data" says which entry it means; and the row and column matrices of 1s (6.4) are just multiplication, used the way the exercises use it. None of the three is an extra topic. Inverse matrices, determinants and the transpose are not in this syllabus. N9 has four bullets and none of them is any of those; there is no matrix division at all. The symbols $M^{- 1}$ , $M^{T}$ and det M do appear in K310's Mathematical Notation appendix, but that list is introduced as "primarily directed towards A-Level", applying to other levels only "where relevant" — the very same list carries complex numbers, which are certainly not examinable here. The content table is what defines the course. Where the textbook raises inverses at all it does so in a Challenge Yourself question and an information box, outside the exercises; the boundary is drawn in 6.3. Transformation matrices and matrix encoding, which the 4A chapter offers as a Journal Writing prompt and an Investigation, are outside N9 for the same reason and are not taught here.

📐 Matrices — Revision Notes

What this chapter is ✕ Not in G2

Order, rows, columns and elements

The four names worth knowing

Reading a matrix — what the entries mean

When are two matrices equal?

Check yourself

Addition and subtraction — and the one thing that can stop them

Multiplying a matrix by a scalar

Matrix equations with unknowns in them

Sums and scalars in a real context

Check yourself

Before you multiply: does the product even exist?

How the multiplication actually works

`AB` is not `BA`

Check yourself

Quantities times prices — and why the shapes have to line up

Past-paper questions

Rows and columns of 1s — totals for free

Two products, one after the other

Check yourself

Past-paper questions

The whole toolkit, on one page ✕ Not in G2

Which operation, and what it needs ✕ Not in G2

The habits that earn the marks

Syllabus at a glance

Ask CherAI

📐 Matrices — Revision Notes

What this chapter is ✕ Not in G2

Order, rows, columns and elements

The four names worth knowing

Reading a matrix — what the entries mean

When are two matrices equal?

Check yourself

Addition and subtraction — and the one thing that can stop them

Multiplying a matrix by a scalar

Matrix equations with unknowns in them

Sums and scalars in a real context

Check yourself

Before you multiply: does the product even exist?

How the multiplication actually works

AB is not BA

Check yourself

Quantities times prices — and why the shapes have to line up

Past-paper questions

Rows and columns of 1s — totals for free

Two products, one after the other

Check yourself

Past-paper questions

The whole toolkit, on one page ✕ Not in G2

Which operation, and what it needs ✕ Not in G2

The habits that earn the marks

Syllabus at a glance

`AB` is not `BA`