What is the determinant of a 2x2 matrix?

For A = pmatrix a & b \\ c & d pmatrix,

What is wrong dimension check?

AB exists only if the columns of A match the rows of B; check before multiplying.

What are cofactor sign slips?

The middle term of a first-row 33 expansion carries a minus sign; the checkerboard pattern is easy to drop.

Compute pmatrix 5 & 2 \\ 3 & 4 pmatrix. [1 mark]

State whether AB = BA holds for general matrices and why. [1 mark]

What does A = 0 tell you about A? [2 marks]

SEAB Original-style practice: Given A = pmatrix 2 & 1 \\ 0 & 3 pmatrix and B = pmatrix 1 & -1 \\ 4 & 2 pmatrix, find AB and BA, and comment on whether they are equal.

Multiply row by column. For AB: $AB = pmatrix 2 & 1 \\ 0 & 3 pmatrixpmatrix 1 & -1 \\ 4 & 2 pmatrix = pmatrix 2(1)+1(4) & 2(-1)+1(2) \\ 0(1)+3(4) & 0(-1)+3(2) pmatrix = pmatrix 6 & 0 \\ 12 & 6 pmatrix.$ For BA: $BA = pmatrix 1 & -1 \\ 4 & 2 pmatrixpmatrix 2 & 1 \\ 0 & 3 pmatrix = pmatrix 1(2)+(-1)(0) & 1(1)+(-1)(3) \\ 4(2)+2(0) & 4(1)+2(3) pmatrix = pmatrix 2 & -2 \\ 8 & 10 pmatrix.$ Since AB ≠ BA, matrix multiplication is not commutative in general. Markers reward correct row-by-column products for both, and the explicit comment that the two products differ so multiplication is non-commutative.

SEAB Original-style practice: Evaluate the determinant of M = pmatrix 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 pmatrix by cofactor expansion along the first row.

Expand along the first row, alternating signs +, -, +: $M = 1vmatrix 4 & 5 \\ 0 & 6 vmatrix - 2vmatrix 0 & 5 \\ 1 & 6 vmatrix + 3vmatrix 0 & 4 \\ 1 & 0 vmatrix.$ Each 22 minor: vmatrix 4 & 5 \\ 0 & 6 vmatrix = 24 - 0 = 24; vmatrix 0 & 5 \\ 1 & 6 vmatrix = 0 - 5 = -5; vmatrix 0 & 4 \\ 1 & 0 vmatrix = 0 - 4 = -4. $M = 1(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22.$ Markers reward the cofactor signs, correct 22 minors, and the arithmetic giving 22.

SingaporeFurther MathsSyllabus dot point

How do we perform matrix arithmetic and compute determinants, and what does a determinant tell us?

Carry out matrix addition and multiplication and evaluate the determinant of 2x2 and 3x3 matrices, interpreting its geometric meaning

A focused answer to the H2 Further Mathematics outcome on matrix arithmetic and determinants. Matrix addition and multiplication, non-commutativity, the determinant of 2x2 and 3x3 matrices by cofactor expansion, its properties, and its geometric meaning as an area or volume scale factor.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to perform matrix addition and multiplication confidently, including recognising when multiplication is defined and that it is not commutative, and to evaluate the determinant of $2\times2$ and $3\times3$ matrices. You should also understand the determinant as a scale factor for area or volume and know its key properties.

The answer

Matrix addition and scalar multiplication

Matrices of the same size are added entry by entry, and a scalar multiplies every entry. These operations are commutative and associative just like ordinary addition.

Matrix multiplication

The product $\mathbf{AB}$ is defined only when the number of columns of $\mathbf{A}$ equals the number of rows of $\mathbf{B}$ . The $(i, j)$ entry of $\mathbf{AB}$ is the dot product of row $i$ of $\mathbf{A}$ with column $j$ of $\mathbf{B}$ :

(\mathbf{AB})_{ij} = \sum_{k} a_{ik}\,b_{kj}.

Multiplication is associative, $\mathbf{(AB)C} = \mathbf{A(BC)}$ , but not commutative: in general $\mathbf{AB} \neq \mathbf{BA}$ . The identity matrix $\mathbf{I}$ satisfies $\mathbf{AI} = \mathbf{IA} = \mathbf{A}$ .

The determinant of a 2x2 matrix

For $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ ,

\det\mathbf{A} = ad - bc.

The determinant of a 3x3 matrix

Expand along any row or column using cofactors. Along the first row of $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ :

\det = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}.

The signs follow the checkerboard pattern $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ . Choosing a row or column with zeros minimises the arithmetic.

Properties and geometric meaning

Key properties: $\det(\mathbf{AB}) = \det\mathbf{A}\,\det\mathbf{B}$ ; $\det(\mathbf{A}^{\mathsf T}) = \det\mathbf{A}$ ; swapping two rows changes the sign; a matrix with a zero row or two equal rows has determinant $0$ . Geometrically, $|\det|$ is the factor by which the linear map scales area (in 2D) or volume (in 3D), and $\det = 0$ means the map collapses space to a lower dimension, so the matrix is singular (non-invertible).

Worked example

Given $\mathbf{A} = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$ , verify that $\det(\mathbf{A}^2) = (\det\mathbf{A})^2$ .

Step 1: Compute det A

\det\mathbf{A} = (3)(4) - (1)(2) = 12 - 2 = 10.

Step 2: Compute A squared

\mathbf{A}^2 = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 9+2 & 3+4 \\ 6+8 & 2+16 \end{pmatrix} = \begin{pmatrix} 11 & 7 \\ 14 & 18 \end{pmatrix}.

Step 3: Compute det of A squared

\det(\mathbf{A}^2) = (11)(18) - (7)(14) = 198 - 98 = 100.

Step 4: Compare

$(\det\mathbf{A})^2 = 10^2 = 100 = \det(\mathbf{A}^2)$ , confirming the multiplicative property $\det(\mathbf{A}^2) = \det\mathbf{A}\cdot\det\mathbf{A}$ .

Common traps

Assuming commutativity: $\mathbf{AB}$ and $\mathbf{BA}$ are usually different and may not both be defined; never swap factors.
Wrong dimension check: $\mathbf{AB}$ exists only if the columns of $\mathbf{A}$ match the rows of $\mathbf{B}$ ; check before multiplying.
Cofactor sign slips: The middle term of a first-row $3\times3$ expansion carries a minus sign; the checkerboard pattern is easy to drop.
Confusing $\det(\mathbf{A} + \mathbf{B})$ with $\det\mathbf{A} + \det\mathbf{B}$: The determinant is not additive; only the multiplicative rule $\det(\mathbf{AB}) = \det\mathbf{A}\det\mathbf{B}$ holds.
Treating a singular matrix as invertible: If $\det = 0$ there is no inverse; do not proceed to compute one.

Examples in context

Example 1. Composing transformations. A rotation followed by a reflection in the plane is the matrix product of their two $2\times2$ matrices, applied right to left. Because multiplication is not commutative, reflecting then rotating generally gives a different map, matching the geometry.

Example 2. Detecting collinearity. Three points in the plane are collinear exactly when a $3\times3$ determinant built from their coordinates is zero, because the enclosed area, which the determinant measures, vanishes.

Try this

Q1. Compute $\det\begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ . [1 mark]

Cue. $5\times4 - 2\times3 = 20 - 6 = 14$ .

Q2. State whether $\mathbf{AB} = \mathbf{BA}$ holds for general matrices and why. [1 mark]

Cue. No; matrix multiplication is not commutative, and the two products may even have different sizes.

Q3. What does $\det\mathbf{A} = 0$ tell you about $\mathbf{A}$ ? [2 marks]

Cue. $\mathbf{A}$ is singular: it has no inverse and its columns are linearly dependent.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksGiven

\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}

and

\mathbf{B} = \begin{pmatrix} 1 & -1 \\ 4 & 2 \end{pmatrix}

, find

\mathbf{AB}

and

\mathbf{BA}

, and comment on whether they are equal.

Show worked answer →

Multiply row by column. For $\mathbf{AB}$ :

\mathbf{AB} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 2(1)+1(4) & 2(-1)+1(2) \\ 0(1)+3(4) & 0(-1)+3(2) \end{pmatrix} = \begin{pmatrix} 6 & 0 \\ 12 & 6 \end{pmatrix}.

For

\mathbf{BA}

\mathbf{BA} = \begin{pmatrix} 1 & -1 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1(2)+(-1)(0) & 1(1)+(-1)(3) \\ 4(2)+2(0) & 4(1)+2(3) \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 8 & 10 \end{pmatrix}.

Since

\mathbf{AB} \neq \mathbf{BA}

, matrix multiplication is not commutative in general.

Markers reward correct row-by-column products for both, and the explicit comment that the two products differ so multiplication is non-commutative.

Original5 marksEvaluate the determinant of

\mathbf{M} = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}

by cofactor expansion along the first row.

Show worked answer →

Expand along the first row, alternating signs $+, -, +$ :

\det\mathbf{M} = 1\begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} - 2\begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} + 3\begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix}.

Each

2\times2

minor:

\begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = 24 - 0 = 24

;

\begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = 0 - 5 = -5

;

\begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = 0 - 4 = -4

\det\mathbf{M} = 1(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22.

Markers reward the cofactor signs, correct $2\times2$ minors, and the arithmetic giving $22$ .