What is a matrix? Notation and order
A matrix is numbers arranged in a rectangle of rows and columns, written inside brackets. We name matrices with capital letters like A, B, C.
- Rows go across. Columns go up and down.
- A matrix with m rows and n columns has order m × n. It holds m·n numbers.
- The number in row i and column j is called aᵢⱼ. We write A = [aᵢⱼ]ₘₓₙ.
Example: A = [5 −2 7; 0 3 1] has order 2 × 3, and a₁₃ = 7. See step 1 of the 3D.
Equality of matrices
Two matrices are equal only if (1) they have the same order and (2) every matching entry is equal. So [x 2; 5 y] = [4 2; 5 1] tells us x = 4 and y = 1. Board questions often use this to make small equations.
Types of matrices
- Row matrix: only one row (order 1 × n).
- Column matrix: only one column (order m × 1).
- Square matrix: rows = columns (n × n). Its entries a₁₁, a₂₂, … form the main diagonal.
- Diagonal matrix: a square matrix with 0 everywhere except the main diagonal.
- Scalar matrix: a diagonal matrix with all diagonal entries equal, like [3 0; 0 3].
Zero matrix and identity matrix
The zero (null) matrix O has every entry 0. Adding O changes nothing: A + O = A.
The identity matrix I is a square matrix with 1 on the main diagonal and 0 elsewhere. Multiplying by I changes nothing: AI = IA = A. It works like the number 1. Step 2 of the 3D shows O (all flat) turning into I (diagonal 1s).
Transpose of a matrix
The transpose Aᵀ (also A′) is made by turning rows into columns. The entry at (i, j) moves to (j, i). An m × n matrix becomes n × m.
Rules: (Aᵀ)ᵀ = A, (A + B)ᵀ = Aᵀ + Bᵀ, (kA)ᵀ = kAᵀ, and (AB)ᵀ = BᵀAᵀ (the order flips, like taking off socks and shoes).
Symmetric and skew-symmetric matrices
A square matrix is symmetric if Aᵀ = A, so aᵢⱼ = aⱼᵢ. It looks like a mirror image across the main diagonal.
It is skew-symmetric if Aᵀ = −A, so aᵢⱼ = −aⱼᵢ. Then every diagonal entry must be 0, because aᵢᵢ = −aᵢᵢ gives aᵢᵢ = 0.
Key fact: every square matrix is a sum of one symmetric and one skew-symmetric matrix: A = ½(A + Aᵀ) + ½(A − Aᵀ).
Addition, subtraction and scalar multiple
Add (or subtract) two matrices of the same order by adding the matching entries. A + B = B + A and (A + B) + C = A + (B + C).
A scalar multiple kA multiplies every entry by the number k. Example: 3[1 −2; 0 4] = [3 −6; 0 12].
Step 4 of the 3D shows the answer blocks growing to the sum of the matching blocks.
Multiplication of matrices
AB is defined only when columns of A = rows of B. If A is m × n and B is n × p, then AB is m × p.
To get cᵢⱼ: walk along row i of A and down column j of B, multiply the pairs, then add.
Example: [2 1; 3 4][1 0; 2 1]: c₁₁ = 2×1 + 1×2 = 4, c₁₂ = 2×0 + 1×1 = 1, c₂₁ = 3×1 + 4×2 = 11, c₂₂ = 3×0 + 4×1 = 4. So AB = [4 1; 11 4]. Step 5 of the 3D grows each block while the row and column glow.
Rules that still work: A(BC) = (AB)C, A(B + C) = AB + AC, AI = IA = A.
Non-commutativity: AB ≠ BA
With numbers, 2 × 3 = 3 × 2. With matrices, usually AB ≠ BA. Using the same A and B as above, BA = [2 1; 7 6], which is not [4 1; 11 4].
Sometimes BA is not even defined: a 2 × 3 times a 3 × 1 works, but a 3 × 1 times a 2 × 3 does not.
Also strange: AB = O can happen even when A ≠ O and B ≠ O. So you cannot "cancel" matrices like numbers. And (A + B)² = A² + AB + BA + B², not A² + 2AB + B².
Invertible matrices and the unique inverse
A square matrix A is invertible if there is a matrix B of the same order with AB = BA = I. Then B is called the inverse, A⁻¹.
Why only one inverse? Suppose B and C both work. Then B = BI = B(AC) = (BA)C = IC = C. So B and C are the same matrix. The inverse is unique.
For a 2 × 2 matrix [a b; c d], if ad − bc ≠ 0: A⁻¹ = (1/(ad − bc))·[d −b; −c a]. If ad − bc = 0, there is no inverse (you will learn why in the Determinants lesson). Also (AB)⁻¹ = B⁻¹A⁻¹.
Try it: the canteen bill
Write how many cups of tea and biscuits your family used on 2 days as a 2 × 2 matrix Q, and their prices as a column P = [10; 5]. Work out QP by hand. Then in the 3D free play, type the same numbers into A and B (put the prices in the first column of B and 0s in the second) and tap AB. Does your answer match the first column?
Exam tip: Matrices and Determinants together carry about 10 marks. Expect 1-mark MCQs on order and types, and 2–3 mark questions on products, transposes and symmetric parts.
Key formulas and definitions
- Order m × n: m rows, n columns, m·n entries; A = [aᵢⱼ]
- A = B ⇔ same order and aᵢⱼ = bᵢⱼ for every i, j
- (A + B)ᵢⱼ = aᵢⱼ + bᵢⱼ (same order only); (kA)ᵢⱼ = k·aᵢⱼ
- (AB)ᵢⱼ = aᵢ₁b₁ⱼ + aᵢ₂b₂ⱼ + … + aᵢₙbₙⱼ; (m × n)(n × p) = m × p
- AI = IA = A, A + O = A, usually AB ≠ BA
- (Aᵀ)ᵀ = A, (A + B)ᵀ = Aᵀ + Bᵀ, (kA)ᵀ = kAᵀ, (AB)ᵀ = BᵀAᵀ
- Symmetric: Aᵀ = A; skew-symmetric: Aᵀ = −A (diagonal all 0)
- A = ½(A + Aᵀ) + ½(A − Aᵀ)
- AB = BA = I ⇒ B = A⁻¹ (unique); (AB)⁻¹ = B⁻¹A⁻¹
- 2 × 2: [a b; c d]⁻¹ = (1/(ad − bc))[d −b; −c a], ad − bc ≠ 0
Worked examples
1. For A = [5 −2 7; 0 3 1], write the order, a₁₃, a₂₂ and the number of entries.
2 rows and 3 columns, so the order is 2 × 3. a₁₃ is row 1, column 3 = 7. a₂₂ = 3. Entries = 2 × 3 = 6.
2. Build the 2 × 2 matrix with aᵢⱼ = i + 2j.
a₁₁ = 1 + 2 = 3, a₁₂ = 1 + 4 = 5, a₂₁ = 2 + 2 = 4, a₂₂ = 2 + 4 = 6. So A = [3 5; 4 6].
3. Find x and y if [x + y 2; 5 x − y] = [6 2; 5 2].
Equal matrices have equal matching entries: x + y = 6 and x − y = 2. Adding: 2x = 8, x = 4. Then y = 2.
4. A = [1 2; 3 4], B = [2 0; 1 −1]. Find A + B and 2A − B.
A + B = [1+2 2+0; 3+1 4−1] = [3 2; 4 3]. 2A = [2 4; 6 8], so 2A − B = [2−2 4−0; 6−1 8+1] = [0 4; 5 9].
5. A = [1 2; 0 1], B = [1 0; 3 1]. Find AB and BA. Are they equal?
AB: c₁₁ = 1·1 + 2·3 = 7, c₁₂ = 1·0 + 2·1 = 2, c₂₁ = 0·1 + 1·3 = 3, c₂₂ = 0·0 + 1·1 = 1 → AB = [7 2; 3 1]. BA: c₁₁ = 1, c₁₂ = 2, c₂₁ = 3·1 + 1·0 = 3, c₂₂ = 3·2 + 1·1 = 7 → BA = [1 2; 3 7]. Not equal, so AB ≠ BA.
6. Write A = [1 4; 2 3] as the sum of a symmetric and a skew-symmetric matrix.
Aᵀ = [1 2; 4 3]. P = ½(A + Aᵀ) = ½[2 6; 6 6] = [1 3; 3 3] (symmetric, Pᵀ = P). Q = ½(A − Aᵀ) = ½[0 2; −2 0] = [0 1; −1 0] (skew, Qᵀ = −Q). Check: P + Q = [1 4; 2 3] = A ✓.
7. Show that B = [3 −1; −5 2] is the inverse of A = [2 1; 5 3].
AB = [2·3 + 1·(−5) 2·(−1) + 1·2; 5·3 + 3·(−5) 5·(−1) + 3·2] = [1 0; 0 1] = I. BA = [3·2 − 1·5 3·1 − 1·3; −5·2 + 2·5 −5·1 + 2·3] = [1 0; 0 1] = I. Both give I, so B = A⁻¹.
8. For A = [3 1; −1 2], show A² − 5A + 7I = O and use it to find A⁻¹.
A² = [9 − 1 3 + 2; −3 − 2 −1 + 4] = [8 5; −5 3]. 5A = [15 5; −5 10]. A² − 5A = [−7 0; 0 −7] = −7I, so A² − 5A + 7I = O ✓. Multiply by A⁻¹: A − 5I + 7A⁻¹ = O, so A⁻¹ = (5I − A)/7 = (1/7)[2 −1; 1 3].
Common mistakes
- Writing the order as columns × rows. It is always rows × columns.
- Multiplying matching entries (like addition) instead of row × column.
- Assuming AB = BA, or using (A + B)² = A² + 2AB + B² for matrices.
- Writing (AB)ᵀ = AᵀBᵀ or (AB)⁻¹ = A⁻¹B⁻¹; the order must flip.