Matrix Operations
A matrix is a rectangular array of numbers. An m×n matrix has m rows and n columns. In 9231, you work primarily with 2×2 and 3×3 square matrices.
Addition and Scalar Multiplication
These are elementwise — only matrices of the same size can be added.
Addition (same dimensions only)
[[a, b], [c, d]] + [[e, f], [g, h]] = [[a+e, b+f], [c+g, d+h]]
Scalar multiplication
k · [[a, b], [c, d]] = [[ka, kb], [kc, kd]]
Matrix Multiplication
The row × column rule: the (i,j) entry of AB is the dot product of row i of A with column j of B. A must be m×n and B must be n×p — the inner dimensions must match.
2×2 multiplication — memorise this pattern
[[a, b], [c, d]] · [[e, f], [g, h]] = [[ae+bg, af+bh], [ce+dg, cf+dh]]
⛔ Matrix Multiplication is NOT Commutative
In general, AB ≠ BA. Always preserve the order. When proving Mᵏ⁺¹ = Mᵏ · M in induction, write it this way — do not swap the order.
Special Matrices
Identity Matrix I
I₂ = [[1,0],[0,1]] I₃ = [[1,0,0],[0,1,0],[0,0,1]]
AI = IA = A for any compatible matrix A.
AI = IA = A for any compatible matrix A.
Zero Matrix O
All entries zero. A + O = A. AO = OA = O.
Note: AB = O does not imply A = O or B = O (unlike real numbers).
Note: AB = O does not imply A = O or B = O (unlike real numbers).
Transpose Aᵀ
Rows become columns: (Aᵀ)ᵢⱼ = Aⱼᵢ
Key property: (AB)ᵀ = BᵀAᵀ — note the reversal of order.
Key property: (AB)ᵀ = BᵀAᵀ — note the reversal of order.
Key Algebraic Properties
| Property | Holds? | Note |
|---|---|---|
| A(B+C) = AB + AC | ✓ Yes | Distributivity holds |
| AB = BA | ✗ Generally No | Commutativity fails |
| (AB)C = A(BC) | ✓ Yes | Associativity holds |
| AB = AC ⟹ B = C | ✗ Not always | Cancellation fails if det(A) = 0 |
| (AB)⁻¹ = B⁻¹A⁻¹ | ✓ Yes | Order reverses — crucial for solving |
| (Aⁿ)ᵀ = (Aᵀ)ⁿ | ✓ Yes | Powers and transpose commute |
Determinants
The determinant is a scalar that encodes key geometric and algebraic information about a matrix. A matrix is invertible if and only if its determinant is non-zero.
2×2 Determinant
For A = [[a,b],[c,d]]
det(A) = |A| = ad − bc
Mnemonic: "down-left diagonal minus up-right diagonal"
3×3 Determinant — Cofactor Expansion
Expand along any row or column. Expanding along row 1 is standard:
Expansion along row 1
det [[a,b,c],[d,e,f],[g,h,k]] = a(ek−fh) − b(dk−fg) + c(dh−eg)
The signs follow the pattern: + − + for row 1. The minor of each element is the 2×2 determinant of the remaining rows and columns.
💡 Sign Pattern for Cofactors
The cofactor Cᵢⱼ = (−1)ⁱ⁺ʲ × (minor). The sign grid for a 3×3 matrix is:
+ − +
− + −
+ − +
− + −
+ − +
Choose the row/column with the most zeros — expansion becomes trivial.
Key Properties of Determinants
| Property | Formula |
|---|---|
| Product rule | det(AB) = det(A) · det(B) |
| Transpose | det(Aᵀ) = det(A) |
| Scalar multiple | det(kA) = kⁿ det(A) for n×n matrix |
| Inverse | det(A⁻¹) = 1 / det(A) |
| Power | det(Aⁿ) = [det(A)]ⁿ |
| Row swap | Swapping two rows negates the determinant |
| Identical rows | det = 0 if any two rows (or columns) are identical |
| Singular matrix | det(A) = 0 ⟺ A is singular (not invertible) |
E
3×3 Determinant — Full Calculation
Find det(A) where A = [[2, 1, −1],[3, 0, 2],[1, −2, 4]].
Expand row 1
det = 2·|[[0,2],[−2,4]]| − 1·|[[3,2],[1,4]]| + (−1)·|[[3,0],[1,−2]]|
Compute minors
|[[0,2],[−2,4]]| = 0·4 − 2·(−2) = 4
|[[3,2],[1,4]]| = 3·4 − 2·1 = 10
|[[3,0],[1,−2]]| = 3·(−2) − 0·1 = −6
|[[3,2],[1,4]]| = 3·4 − 2·1 = 10
|[[3,0],[1,−2]]| = 3·(−2) − 0·1 = −6
Combine
det = 2(4) − 1(10) + (−1)(−6) = 8 − 10 + 6
det(A) = 4
Matrix Inverses
The inverse A⁻¹ satisfies AA⁻¹ = A⁻¹A = I. It exists if and only if det(A) ≠ 0.
2×2 Inverse — Direct Formula
For A = [[a,b],[c,d]] with det(A) = ad−bc ≠ 0:
2×2 inverse formula — must memorise
A⁻¹ = (1/det(A)) · [[d, −b],[−c, a]]
Swap the diagonal entries (a,d) and negate the off-diagonal entries (b,c). Then divide by det(A).
3×3 Inverse — Adjugate Method
Step 1 — Det
Compute det(A). If det(A) = 0, the matrix is singular and no inverse exists. State this and stop.
Step 2 — Minors
Find the 9 minors Mᵢⱼ — the 2×2 determinant obtained by deleting row i and column j.
Step 3 — Cofactors
Apply signs: Cᵢⱼ = (−1)ⁱ⁺ʲ · Mᵢⱼ. This gives the cofactor matrix C.
Step 4 — Adjugate
adj(A) = Cᵀ — transpose the cofactor matrix. Each cofactor moves to its mirror position.
Step 5 — Inverse
A⁻¹ = adj(A) / det(A). Divide every entry of adj(A) by det(A).
⚠ Always verify: check A · A⁻¹ = I for at least one row.
⚠ Always verify: check A · A⁻¹ = I for at least one row.
E
Full 3×3 Inverse — Step by Step
Find A⁻¹ where A = [[2, 1, 0],[1, 3, 1],[0, 1, 2]].
Step 1 — det
det = 2(3·2−1·1) − 1(1·2−1·0) + 0
= 2(5) − 1(2) = 10 − 2 = 8
= 2(5) − 1(2) = 10 − 2 = 8
Step 2&3 — Cofactors
C₁₁ = +(3·2−1·1) = 5 C₁₂ = −(1·2−1·0) = −2 C₁₃ = +(1·1−3·0) = 1
C₂₁ = −(1·2−0·1) = −2 C₂₂ = +(2·2−0·0) = 4 C₂₃ = −(2·1−1·0) = −2
C₃₁ = +(1·1−0·3) = 1 C₃₂ = −(2·1−0·1) = −2 C₃₃ = +(2·3−1·1) = 5
C₂₁ = −(1·2−0·1) = −2 C₂₂ = +(2·2−0·0) = 4 C₂₃ = −(2·1−1·0) = −2
C₃₁ = +(1·1−0·3) = 1 C₃₂ = −(2·1−0·1) = −2 C₃₃ = +(2·3−1·1) = 5
Step 4 — adj(A)
Transpose C:
adj(A) = [[5,−2,1],[−2,4,−2],[1,−2,5]]
adj(A) = [[5,−2,1],[−2,4,−2],[1,−2,5]]
Step 5 — A⁻¹
A⁻¹ = (1/8)·[[5,−2,1],[−2,4,−2],[1,−2,5]]
= [[5/8, −1/4, 1/8],[−1/4, 1/2, −1/4],[1/8, −1/4, 5/8]]
Verify row 1
Row 1 of A · Col 1 of A⁻¹: 2·(5/8) + 1·(−1/4) + 0·(1/8) = 10/8 − 2/8 = 1 ✓
Important Inverse Properties
Product inverse
(AB)⁻¹ = B⁻¹A⁻¹
Order reverses — same as (AB)ᵀ = BᵀAᵀ
Power inverse
(Aⁿ)⁻¹ = (A⁻¹)ⁿ
Used when solving matrix equations involving powers
Determinant of inverse
det(A⁻¹) = 1 / det(A)
Double inverse
(A⁻¹)⁻¹ = A
Solving Linear Systems
A system of n linear equations in n unknowns can be written as Ax = b where A is the coefficient matrix, x is the column of unknowns, and b is the column of constants.
Matrix form of 2×2 system
[[a, b],[c, d]] · [[x],[y]] = [[p],[q]]
⟹ Ax = b
Solution by Inverse
If det(A) ≠ 0
x = A⁻¹b
Pre-multiply both sides by A⁻¹. This gives a unique solution.
Three Cases — Determined by det(A)
| det(A) | Geometric Interpretation (3×3) | Number of Solutions |
|---|---|---|
| ≠ 0 | Three planes meet at a single point | Unique solution |
| = 0 | Planes meet along a line (sheaf) or are consistent but not independent | Infinitely many solutions |
| = 0 | Planes form a triangular prism, or two/three are parallel | No solution (inconsistent) |
⚠️ Cambridge Examination Point — Singular Systems
When det(A) = 0, Cambridge asks you to determine whether the system is consistent (infinitely many solutions) or inconsistent (no solution). The method is to use row reduction or to substitute and check for contradiction. A common question gives a parameter λ and asks for the value making the system singular, then investigates consistency at that λ.
E
Parametric System — Find λ for Singularity, then Investigate
The system Ax = b has A = [[1, 2, 1],[2, λ, 3],[1, 1, 2]] and b = [[4],[7],[3]].
(i) Find the value of λ for which the system does not have a unique solution.
(ii) For this value of λ, determine whether solutions exist.
(i) Find the value of λ for which the system does not have a unique solution.
(ii) For this value of λ, determine whether solutions exist.
Part (i)
det(A) = 1(2λ−3) − 2(4−3) + 1(2−λ)
= 2λ−3 − 2 + 2−λ = λ − 3
det(A) = 0 when
= 2λ−3 − 2 + 2−λ = λ − 3
det(A) = 0 when
λ = 3
Part (ii) λ=3
System: x+2y+z=4, 2x+3y+3z=7, x+y+2z=3
Row 2 − 2×Row 1: −y+z = −1
Row 3 − Row 1: −y+z = −1 ✓ (same equation — consistent)
Let z=t: y=t+1, x=4−2(t+1)−t=2−3t
Row 2 − 2×Row 1: −y+z = −1
Row 3 − Row 1: −y+z = −1 ✓ (same equation — consistent)
Let z=t: y=t+1, x=4−2(t+1)−t=2−3t
Infinitely many solutions: (2−3t, 1+t, t) for t ∈ ℝ
Linear Transformations
A 2×2 matrix M represents a linear transformation of ℝ². The image of a point (x, y) is found by multiplying: M · [[x],[y]].
Standard 2×2 Transformation Matrices
| Transformation | Matrix | det | Effect on Area |
|---|---|---|---|
| Reflection in x-axis | [[1,0],[0,−1]] | −1 | Preserves area, reverses orientation |
| Reflection in y-axis | [[−1,0],[0,1]] | −1 | Preserves area, reverses orientation |
| Reflection in y=x | [[0,1],[1,0]] | −1 | Preserves area, reverses orientation |
| Reflection in y=−x | [[0,−1],[−1,0]] | −1 | Preserves area, reverses orientation |
| Rotation by θ anticlockwise | [[cosθ,−sinθ],[sinθ,cosθ]] | 1 | Preserves area and orientation |
| Enlargement scale factor k | [[k,0],[0,k]] | k² | Area scaled by k² |
| Stretch parallel to x-axis | [[k,0],[0,1]] | k | Area scaled by k |
| Shear parallel to x-axis | [[1,k],[0,1]] | 1 | Preserves area |
📌 Key Geometric Facts
- |det(M)| gives the scale factor for areas. A unit square maps to a parallelogram of area |det(M)|.
- det(M) < 0 means the transformation includes a reflection (orientation is reversed).
- det(M) = 0 means the transformation collapses 2D space to a line or point — not invertible.
- To find the image of a shape, transform each vertex and then join the images.
- To find the transformation given image and pre-image, use the two equations M·v₁ = w₁ and M·v₂ = w₂.
Composition of Transformations
Applying transformation A followed by transformation B gives the combined matrix BA (note order — B is applied second but written first).
Order matters
"First apply A, then B" ⟹ combined matrix = BA
Remember: matrix multiplication acts right-to-left on column vectors. The rightmost matrix acts first.
E
Finding the Transformation Matrix
A linear transformation T maps (1,0) → (3,1) and (0,1) → (−1,2). Find the matrix of T and the image of the triangle with vertices O(0,0), A(2,0), B(1,3).
Matrix
Columns of M are the images of the standard basis vectors:
M = [[3,−1],[1,2]]
Image of O
O' = (0,0) — origin always maps to itself under linear transformation.
Image of A(2,0)
[[3,−1],[1,2]]·[[2],[0]] = [[6],[2]]. So A' = (6,2).
Image of B(1,3)
[[3,−1],[1,2]]·[[1],[3]] = [[0],[7]]. So B' = (0,7).
Area check
Original area = ½|det([[2,1],[0,3]])| = ½·6 = 3.
det(M) = 6+1 = 7. Image area = 3·7 = 21. ✓
det(M) = 6+1 = 7. Image area = 3·7 = 21. ✓
Worked Examples
1
Matrix Equation — Solve for X
Given A = [[3,1],[5,2]] and B = [[7,2],[11,3]], solve AX = B for the matrix X.
Find A⁻¹
det(A) = 6−5 = 1. A⁻¹ = (1/1)·[[2,−1],[−5,3]] = [[2,−1],[−5,3]]
Solve
AX = B ⟹ X = A⁻¹B
X = [[2,−1],[−5,3]]·[[7,2],[11,3]]
= [[14−11, 4−3],[−35+33, −10+9]]
X = [[2,−1],[−5,3]]·[[7,2],[11,3]]
= [[14−11, 4−3],[−35+33, −10+9]]
X = [[3,1],[−2,−1]]
Verify
AX = [[3,1],[5,2]]·[[3,1],[−2,−1]] = [[9−2,3−1],[15−4,5−2]] = [[7,2],[11,3]] = B ✓
2
Determinant with Parameter
Find the values of k for which the matrix M = [[k,2,0],[1,k,1],[2,0,k]] is singular.
Expand det
det(M) = k(k²−0) − 2(k−2) + 0
= k³ − 2k + 4
= k³ − 2k + 4
Set = 0
k³ − 2k + 4 = 0
Try k = −2: −8+4+4 = 0 ✓
Factor: (k+2)(k²−2k+2) = 0
k²−2k+2 = 0 has discriminant 4−8 = −4 < 0 (no real roots)
Try k = −2: −8+4+4 = 0 ✓
Factor: (k+2)(k²−2k+2) = 0
k²−2k+2 = 0 has discriminant 4−8 = −4 < 0 (no real roots)
k = −2 (only real solution)
3
Transformation — Rotation then Reflection
R is rotation by 90° anticlockwise. S is reflection in the x-axis. Find the single matrix representing "first R, then S" and describe the resulting transformation geometrically.
Matrices
R = [[cos90°, −sin90°],[sin90°, cos90°]] = [[0,−1],[1,0]]
S = [[1,0],[0,−1]]
S = [[1,0],[0,−1]]
Combined = SR
(S applied after R) = SR = [[1,0],[0,−1]]·[[0,−1],[1,0]]
= [[0,−1],[−1,0]]
= [[0,−1],[−1,0]]
SR = [[0,−1],[−1,0]]
Geometry
det(SR) = 0·0 − (−1)(−1) = −1. Orientation reversed → reflection.
Test: (1,0) → (0,−1); (0,1) → (−1,0). This is reflection in y = −x.
Test: (1,0) → (0,−1); (0,1) → (−1,0). This is reflection in y = −x.
Practice Questions
Question 1 — 3×3 Determinant
[3 marks]
Find det(A) where A = [[3,0,−1],[2,1,4],[−1,2,3]].
Expand along row 1. The zero in position (1,2) eliminates one minor entirely — choose expansion carefully.
✓ Solution
Expanding along row 1 (middle term is zero so only two minors):det = 3·|[[1,4],[2,3]]| − 0 + (−1)·|[[2,1],[−1,2]]|
= 3(3−8) − 1(4+1)
= 3(−5) − 5
= −20
Question 2 — 2×2 Inverse and Matrix Equation
[5 marks]
Given P = [[4,3],[3,2]] and Q = [[5],[1]], solve Px = Q. Hence find x and y.
Find P⁻¹ using the 2×2 formula (swap diagonal, negate off-diagonal, divide by det). Then x = P⁻¹Q.
✓ Solution
det(P) = 8−9 = −1P⁻¹ = (1/−1)·[[2,−3],[−3,4]] = [[−2,3],[3,−4]]
x = P⁻¹Q = [[−2,3],[3,−4]]·[[5],[1]] = [[−10+3],[15−4]]
x = [[−7],[11]], so x = −7, y = 11
Verify: 4(−7)+3(11) = −28+33 = 5 ✓ 3(−7)+2(11) = −21+22 = 1 ✓
Question 3 — Singular Matrix, Parameter
[4 marks]
The matrix A = [[2,λ,1],[1,2,λ],[λ,1,2]] is singular. Find the possible values of λ.
Compute det(A) by cofactor expansion along row 1. You will get a cubic in λ. Try λ=1 and λ=−2 as possible rational roots.
✓ Solution
det(A) = 2(4−λ) − λ(2−λ²) + 1(1−2λ)= 8−2λ − 2λ+λ³ + 1−2λ
= λ³ − 6λ + 9... [re-expand carefully]
det = 2(4−λ) − λ(2−λ²) + 1(1−2λ)
= 8−2λ − 2λ+λ³ + 1−2λ
= λ³ − 6λ + 9
Wait — recheck: 8+1=9, −2λ−2λ−2λ=−6λ ✓
Set λ³−6λ+9 = ... try λ=−3: −27+18+9=0 ✓
Factor: (λ+3)(λ²−3λ+3)=0
Discriminant of quadratic: 9−12=−3<0 (no real roots)
λ = −3 (only real solution)
Question 4 — Transformation and Area
[5 marks]
The matrix M = [[2,1],[−1,3]] represents a linear transformation T.
(i) Find the image of the point (3, −1) under T.
(ii) The unit square has vertices O(0,0), A(1,0), B(1,1), C(0,1). Find the area of the image parallelogram.
(i) Find the image of the point (3, −1) under T.
(ii) The unit square has vertices O(0,0), A(1,0), B(1,1), C(0,1). Find the area of the image parallelogram.
For (ii) the area scale factor is |det(M)|.
✓ Solution
(i) M·[[3],[−1]] = [[6−1],[−3−3]] = [[5],[−6]]. Image point: (5, −6).(ii) det(M) = 6−(−1) = 7. Area = |7|×1 =
7 square units
Interactive Matrix Calculator
Enter matrix entries and choose an operation. All steps are shown.
2×2 Matrix Operations
Matrix A
Result
Select an operation above and enter values.
Formula Sheet — Matrices
2×2 Determinant & Inverse
det [[a,b],[c,d]]ad − bc
Inverse (det≠0)(1/det)·[[d,−b],[−c,a]]
Singulardet = 0 ⟺ no inverse
3×3 Determinant
Expand row 1a·M₁₁ − b·M₁₂ + c·M₁₃
Sign pattern+−+ / −+− / +−+
Choose row with most zerosFewer minors to compute
3×3 Inverse
A⁻¹ = adj(A)/det(A)—
Cofactor Cᵢⱼ(−1)ⁱ⁺ʲ · Mᵢⱼ
adj(A)Transpose of cofactor matrix
VerifyAA⁻¹ = I
Key Properties
det(AB)det(A)·det(B)
(AB)⁻¹B⁻¹A⁻¹
(AB)ᵀBᵀAᵀ
det(kA) for n×nkⁿ·det(A)
Linear Systems Ax=b
det≠0Unique solution x=A⁻¹b
det=0, consistent∞ solutions
det=0, inconsistentNo solution
Transformations
Area scale factor|det(M)|
det < 0Includes reflection
Rotation by θ[[cosθ,−sinθ],[sinθ,cosθ]]
Order: A then BCombined = BA
📋 Cambridge Exam Strategy — Matrices
- Determinant with parameter: Expand fully, set = 0, solve the resulting polynomial. Always try simple integer values first (Factor Theorem).
- Singular system: After finding λ, substitute back and row-reduce to determine consistency. Cambridge expects a geometric interpretation (line of intersection, or contradiction).
- 3×3 inverse: If the question says "hence solve", use A⁻¹b directly — do not row-reduce again. Show at least one verification entry.
- Transformation questions: The image of the standard basis vectors are the columns of M — use this to find M quickly.
- Matrix equations: Always pre-multiply or post-multiply by the inverse on the correct side — AB ≠ BA so side matters.
- Area/volume: Area scale = |det|. For 3D (if asked), volume scale = |det| for a 3×3 transformation.