Matrix Multiplication

Matrix multiplication is the single operation that makes matrices powerful. It is not element-by-element — it follows a "row times column" rule that, at first sight, looks arbitrary, but encodes exactly the right thing: the composition of linear maps. Multiplying the matrices of two transformations gives the matrix of "do one, then the other". With that one idea, geometry, simultaneous equations and iterated processes all collapse into algebra. The price is that matrix multiplication behaves differently from ordinary multiplication: it is not commutative, it has zero divisors, and you cannot cancel freely. This lesson develops the rule, the algebra, and the exam technique around all three surprises. By the end you should be able to multiply $2\times2$2×2 and $3\times3$3×3 matrices fluently, decide when a product is defined, compute powers, prove a power formula by induction, and explain each of the three "anomalies" with a worked counter-example — the exact skill set the examiners test.

---

1. Where this sits in AQA 7367

Matrix multiplication is compulsory pure content for Papers 1 and 2. It is heavily AO1 (carry out the multiplication accurately) but the non-commutativity, the zero-product property and proof-by-induction conjectures for $A^n$An are squarely AO2 (reason and communicate). Multiplication is the engine for everything downstream: determinants ($\det(AB) = \det A\,\det B$det(AB)=detAdetB), inverses ($AA^{-1} = I$AA−1=I), linear transformations (composition $=$= product), solving systems ($A\mathbf{x} = \mathbf{b}$Ax=b), and diagonalisation ($A^n = PD^nP^{-1}$An=PDnP−1). Fluency here pays off in every later lesson.

---

2. Core theory: the multiplication rule

To multiply matrices $A$A (order $m \times n$m×n) and $B$B (order $n \times p$n×p):

• the number of columns of $A$A must equal the number of rows of $B$B (the "inner" dimensions match);
• the product $AB$AB has order $m \times p$m×p (the "outer" dimensions);
• the entry in row $i$i, column $j$j of $AB$AB is the dot product of row $i$i of $A$A with column $j$j of $B$B:

$$(AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}.$$(AB)ij​=k=1∑n​aik​bkj​.

A useful slogan: inner dimensions must agree; outer dimensions give the answer's size. So a $2\times3$2×3 times a $3\times4$3×4 gives a $2\times4$2×4, but a $2\times3$2×3 times a $2\times3$2×3 is undefined.

For the $2\times2$2×2 case, written out in full:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}.$$(ac​bd​)(eg​fh​)=(ae+bgce+dg​af+bhcf+dh​).

Read each output entry as "row from the left, column from the right": top-left is (row 1)·(col 1) $= ae + bg$=ae+bg; top-right is (row 1)·(col 2) $= af + bh$=af+bh; and so on.

Worked Example 1 — a 2×2 product (with mark scheme)

$$\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}.$$(21​34​)(50​−12​).

Compute entry by entry:

$$\begin{aligned} (1,1)&: 2\cdot5 + 3\cdot0 = 10, & (1,2)&: 2\cdot(-1) + 3\cdot2 = 4, \\ (2,1)&: 1\cdot5 + 4\cdot0 = 5, & (2,2)&: 1\cdot(-1) + 4\cdot2 = 7. \end{aligned}$$(1,1)(2,1)​:2⋅5+3⋅0=10,:1⋅5+4⋅0=5,​(1,2)(2,2)​:2⋅(−1)+3⋅2=4,:1⋅(−1)+4⋅2=7.​ $$\therefore\; \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 10 & 4 \\ 5 & 7 \end{pmatrix}.$$∴(21​34​)(50​−12​)=(105​47​).

Mark scheme: M1 for a correct method seen on at least one entry (clear row·column dot product); A1 for two correct entries; A1 for the fully correct product. Showing the four dot products explicitly secures the M1 even if one arithmetic slip costs an A1.

Worked Example 2 — a 3×3 product (with mark scheme)

$$\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 3 & 2 & 0 \end{pmatrix}.$$​102​210​031​​​103​012​210​​.

Take row 1 of the left, $(1, 2, 0)$(1,2,0), dotted with each column of the right: $1\cdot1+2\cdot0+0\cdot3 = 1$1⋅1+2⋅0+0⋅3=1; $1\cdot0+2\cdot1+0\cdot2 = 2$1⋅0+2⋅1+0⋅2=2; $1\cdot2+2\cdot1+0\cdot0 = 4$1⋅2+2⋅1+0⋅0=4. Row 2 $(0,1,3)$(0,1,3): $0+0+9 = 9$0+0+9=9; $0+1+6 = 7$0+1+6=7; $0+1+0 = 1$0+1+0=1. Row 3 $(2,0,1)$(2,0,1): $2+0+3 = 5$2+0+3=5; $0+0+2 = 2$0+0+2=2; $4+0+0 = 4$4+0+0=4. Hence

$$= \begin{pmatrix} 1 & 2 & 4 \\ 9 & 7 & 1 \\ 5 & 2 & 4 \end{pmatrix}.$$=​195​272​414​​.

Mark scheme: M1 for correct row·column method on at least one row; A1 for any one fully correct row; A1 for a second correct row; A1 for the complete matrix. Laying the work out row-by-row, as above, makes the method marks unambiguous.

---

3. Matrix multiplication is NOT commutative

In general, $AB \neq BA$AB=BA — and frequently one of them is not even defined. This is the deepest difference from ordinary numbers.

Counter-example. Let $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$A=(10​21​) and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$B=(01​10​). Then

$$AB = \begin{pmatrix} 1\cdot0+2\cdot1 & 1\cdot1+2\cdot0 \\ 0\cdot0+1\cdot1 & 0\cdot1+1\cdot0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}, \qquad BA = \begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix}.$$AB=(1⋅0+2⋅10⋅0+1⋅1​1⋅1+2⋅00⋅1+1⋅0​)=(21​10​),BA=(01​12​).

Since $\begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} \neq \begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix}$(21​10​)=(01​12​), we have $AB \neq BA$AB=BA. Geometrically this is obvious: a shear followed by a swap of axes is not the same as a swap followed by a shear. Two matrices that do satisfy $AB = BA$AB=BA are called commuting — for instance any matrix commutes with $I$I, with the zero matrix, and with its own powers; diagonal matrices commute with one another.

There is an even sharper way non-commutativity shows up. For numbers, $(a+b)^2 = a^2 + 2ab + b^2$(a+b)2=a2+2ab+b2. For matrices,

$$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2,$$(A+B)2=(A+B)(A+B)=A2+AB+BA+B2,

and the middle two terms collapse to $2AB$2AB only if $A$A and $B$B commute. In general $AB + BA \neq 2AB$AB+BA=2AB, so the familiar binomial expansion silently fails — a classic source of lost marks. Always expand a bracketed matrix square the long way unless you have proved the factors commute.

Exam Tip: Never silently swap the order of a matrix product. Where a question asks for $AB$AB, compute $AB$AB, not $BA$BA. The order is examined precisely because it is a common slip.

---

4. Specimen-style exam question

(Specimen-style — not a real past paper.) The matrices $A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$A=(21​−13​) and $B = \begin{pmatrix} 1 & 0 \\ 4 & -2 \end{pmatrix}$B=(14​0−2​) are given. (a) Find $AB$AB and $BA$BA. (b) Hence verify that $AB - BA$AB−BA has trace $0$0. (5 marks)

Model solution.

$$AB = \begin{pmatrix} 2\cdot1+(-1)\cdot4 & 2\cdot0+(-1)\cdot(-2) \\ 1\cdot1+3\cdot4 & 1\cdot0+3\cdot(-2) \end{pmatrix} = \begin{pmatrix} -2 & 2 \\ 13 & -6 \end{pmatrix},$$AB=(2⋅1+(−1)⋅41⋅1+3⋅4​2⋅0+(−1)⋅(−2)1⋅0+3⋅(−2)​)=(−213​2−6​), $$BA = \begin{pmatrix} 1\cdot2+0\cdot1 & 1\cdot(-1)+0\cdot3 \\ 4\cdot2+(-2)\cdot1 & 4\cdot(-1)+(-2)\cdot3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 6 & -10 \end{pmatrix}.$$BA=(1⋅2+0⋅14⋅2+(−2)⋅1​1⋅(−1)+0⋅34⋅(−1)+(−2)⋅3​)=(26​−1−10​).

Then $AB - BA = \begin{pmatrix} -4 & 3 \\ 7 & 4 \end{pmatrix}$AB−BA=(−47​34​), whose trace is $-4 + 4 = 0$−4+4=0. (M1 method for a product; A1 $AB$AB; A1 $BA$BA; M1 forming $AB-BA$AB−BA; A1 trace $=0$=0.)

This is no coincidence: $\operatorname{tr}(AB) = \operatorname{tr}(BA)$tr(AB)=tr(BA) for all square matrices, so the commutator $AB - BA$AB−BA always has trace zero — a fact worth quoting.

---

5. The algebra of multiplication

Property	Statement
Associative	$(AB)C = A(BC)$(AB)C=A(BC)
Distributive (left)	$A(B + C) = AB + AC$A(B+C)=AB+AC
Distributive (right)	$(A + B)C = AC + BC$(A+B)C=AC+BC
Identity	$AI = IA = A$AI=IA=A
Scalar	$k(AB) = (kA)B = A(kB)$k(AB)=(kA)B=A(kB)
NOT commutative	$AB \neq BA$AB=BA in general
NOT cancellative	$AB = AC$AB=AC does not imply $B = C$B=C

The identity matrix

The identity $I$I plays the role of the number $1$1: $AI = IA = A$AI=IA=A. For $2\times2$2×2, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$I=(10​01​); for $3\times3$3×3, $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$I=​100​010​001​​.

Worked Example (associativity in action)

Associativity $(AB)C = A(BC)$(AB)C=A(BC) lets you bracket a triple product whichever way is easiest. With $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$A=(10​21​), $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$B=(11​01​), $C = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$C=(20​11​):

$$AB = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \implies (AB)C = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 6 & 5 \\ 2 & 2 \end{pmatrix},$$AB=(31​21​)⟹(AB)C=(31​21​)(20​11​)=(62​52​), $$BC = \begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix} \implies A(BC) = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 5 \\ 2 & 2 \end{pmatrix}.$$BC=(22​12​)⟹A(BC)=(10​21​)(22​12​)=(62​52​).

Both bracketings give $\begin{pmatrix} 6 & 5 \\ 2 & 2 \end{pmatrix}$(62​52​), confirming associativity — and showing you may always choose the easier order to compute.

Multiplying a matrix by a vector

A matrix times a column vector returns a column vector — the foundation of linear transformations:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.$$(ac​bd​)(xy​)=(ax+bycx+dy​).

Because $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix}$(ac​bd​)(10​)=(ac​) and $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} b \\ d \end{pmatrix}$(ac​bd​)(01​)=(bd​), the columns of the matrix are precisely the images of the basis vectors — the key idea of the transformation lessons.

A worked instance: $\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 2\cdot5 + (-1)\cdot2 \\ 3\cdot5 + 4\cdot2 \end{pmatrix} = \begin{pmatrix} 8 \\ 23 \end{pmatrix}$(23​−14​)(52​)=(2⋅5+(−1)⋅23⋅5+4⋅2​)=(823​). The vector $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$(52​) is sent to $\begin{pmatrix} 8 \\ 23 \end{pmatrix}$(823​) — this is exactly how a transformation moves a point.

Non-square products

Multiplication is not confined to square matrices. A $2\times3$2×3 times a $3\times1$3×1 gives a $2\times1$2×1:

$$\begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix}\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1\cdot4 + 0\cdot2 + 2\cdot1 \\ -1\cdot4 + 3\cdot2 + 1\cdot1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}.$$(1−1​03​21​)​421​​=(1⋅4+0⋅2+2⋅1−1⋅4+3⋅2+1⋅1​)=(63​).

The reverse product $\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix}$​421​​(1−1​03​21​) is undefined (inner dimensions $1$1 and $2$2 disagree) — another way order matters.

---

6. Matrix powers

For a square matrix $A$A, powers are repeated multiplication:

$A^0 = I, \quad A^1 = A, \quad A^2 = AA, \quad A^3 = A^2A, \quad \ldots, \quad A^{m+n} = A^mA^n.$A0=I,A1=A,A2=AA,A3=A2A,…,Am+n=AmAn.

Worked Example 3 — powers and a conjecture (with mark scheme)

If $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$A=(10​11​), find $A^2$A2 and $A^3$A3, and conjecture $A^n$An.

$$A^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \qquad A^3 = A^2A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}.$$A2=(10​11​)(10​11​)=(10​21​),A3=A2A=(10​21​)(10​11​)=(10​31​).

The pattern suggests $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$An=(10​n1​). This is provable by induction: assuming $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$Ak=(10​k1​), then