9231 Further Pure Mathematics — Paper 1
a b c d M·v original image det M = ad−bc area scale factor invariant line
9231 Further Pure › Paper 1 › Lesson 1.4

Matrices &
Transformations

Operations, determinants, inverses, 2D geometric transformations, area scale factor, invariant points and invariant lines.

🎯 Cambridge 9231 P1 📐 Section 1.4 ⭐ Invariants Key Topic ⏱ ~110 min
1.4 Lesson

1. Matrix Operations

Matrices can have at most 3 rows and 3 columns in this syllabus. The key operations are addition, subtraction, and multiplication — each with strict size conditions.

Addition / Subtraction

Only possible when matrices have the same dimensions. Add corresponding entries.

A + BAdd element by element
Must haveSame number of rows and columns

Multiplication

A is m×n, B is n×p → AB is m×p. The inner dimensions must match.

(AB)ᵢⱼ= row i of A · column j of B
Order mattersAB ≠ BA in general
💡 Special Matrices
Zero matrix 0All entries zero — additive identity
Identity I1s on diagonal, 0s elsewhere — AI = IA = A
Singulardet = 0 — no inverse exists
Non-singulardet ≠ 0 — inverse exists

2. Determinants

2×2 Determinant

det(M) for M = [[a,b],[c,d]]:   det = ad − bc

3×3 Determinant — Cofactor Expansion Along Row 1

For a 3×3 matrix M = [[a,b,c],[d,e,f],[g,h,i]]:

det(M) = a × (ei − fh) − b × (di − fg) + c × (dh − eg)
Sign pattern for cofactors: + − + / − + − / + − +   (checkerboard)
⚠ Sign Pattern — Never Forget

The cofactor signs for a 3×3 matrix follow the checkerboard: top-left is +, then alternates. The entry a has cofactor +(ei−fh), b has −(di−fg), c has +(dh−eg). Always apply this pattern.

3. Inverse Matrices

2×2 Inverse

⭐ 2×2 Inverse Formula
M⁻¹ = (1/det M) × [[d, −b], [−c, a]] swap main diag, negate off-diag
Only if det M ≠ 0 (non-singular)

3×3 Inverse — Matrix of Cofactors Method

Three steps for 3×3 inverse
Step 1Find det(M). If zero, stop — M is singular.
Step 2Find the matrix of cofactors C (each entry: cofactor of that position)
Step 3M⁻¹ = (1/det M) × Cᵀ   (transpose of cofactor matrix)

Key Property: (AB)⁻¹ = B⁻¹A⁻¹

💡 Reversal Rule
(AB)⁻¹= B⁻¹A⁻¹order reverses
(ABC)⁻¹= C⁻¹B⁻¹A⁻¹extends to products of 3+
Verify:(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I ✓

4. 2D Geometric Transformations

A 2×2 matrix M transforms a point (x, y) to M·[x, y]ᵀ. The images of basis vectors (1,0) and (0,1) form the columns of M.

⭐ Column Rule — How to Find a Transformation Matrix

The first column of M = image of (1,0). The second column = image of (0,1). Apply the transformation to these two basis vectors to build the matrix.

Rotation by θ (anticlockwise)
[cosθ −sinθ]
[sinθ cosθ]
det:1
area:unchanged
inv:rotate by −θ
Reflection in y = x
[0 1]
[1 0]
det:−1
area:unchanged (|det|=1)
inv:itself (self-inverse)
Reflection in x-axis
[1 0]
[0 −1]
det:−1
invariant:x-axis
inv:itself
Reflection in y-axis
[−1 0]
[ 0 1]
det:−1
invariant:y-axis
inv:itself
Enlargement, scale k
[k 0]
[0 k]
det:k²
area:× k²
inv:scale 1/k
Stretch in x by factor k
[k 0]
[0 1]
det:k
invariant:y-axis
inv:stretch x by 1/k
Shear in x-direction (factor k)
[1 k]
[0 1]
det:1
invariant:x-axis
area:unchanged
Rotation 90° anticlockwise
[0 −1]
[1 0]
det:1
θ = 90°cos90°=0, sin90°=1
inv:rotate 90° clockwise

Composition of Transformations — Order Matters

⚠ Critical: AB Means B First, Then A

The matrix product AB represents the transformation B applied first, then A applied to the result. This is because we multiply on the left:

image= AB · v = A(B · v)B acts on v first
"B then A"→ matrix product AB (not BA)

Area Scale Factor

Area scale factor= |det M|absolute value of determinant
Sign of det M+ : orientation preserved   − : orientation reversed

5. Invariant Points and Invariant Lines

Invariant Points

A point P is invariant under transformation M if M maps P to itself: M·P = P.

ConditionM·[x,y]ᵀ = [x,y]ᵀ
Gives(M − I)[x,y]ᵀ = 0

The origin (0,0) is always an invariant point for any linear transformation through the origin.

Invariant Lines

A line is invariant if every point on the line maps to a point that is also on the line (not necessarily the same point).

MethodLet point (x, mx) be on line y=mx
Apply MFind image (x', y')
Requirey' = mx' (image on same line)

Method for Finding Invariant Lines Through the Origin

  1. 1Let a general point on the line y = mx be (t, mt) for any scalar t.
  2. 2Apply the matrix M to this point: M·[t, mt]ᵀ = [x', y']ᵀ.
  3. 3Require that the image lies on the same line: y' = m·x'.
  4. 4This gives an equation in m — solve for m to find the gradient(s) of invariant lines.
📐 Invariant Lines NOT Through Origin

For invariant lines y = mx + c (c ≠ 0), let a general point be (t, mt + c) and require the image also satisfies y = mx + c. This gives conditions on both m and c.

Example 1 3×3 Determinant and Inverse ★★☆ Challenging
Find the determinant and inverse of the matrix M = [[2,1,−1],[0,3,2],[1,−1,4]].
1
Expand along row 1
det M= 2(3·4 − 2·(−1)) − 1(0·4 − 2·1) + (−1)(0·(−1) − 3·1)
= 2(12 + 2) − 1(0 − 2) + (−1)(0 − 3)
= 2(14) − 1(−2) + (−1)(−3)
det M= 28 + 2 + 3 = 33
2
Matrix of cofactors C (with checkerboard signs)
C₁₁= +(3·4−2·(−1)) = +14
C₁₂= −(0·4−2·1) = −(−2) = +2
C₁₃= +(0·(−1)−3·1) = −3
C₂₁= −(1·4−(−1)·(−1)) = −(4−1) = −3
C₂₂= +(2·4−(−1)·1) = +(8+1) = +9
C₂₃= −(2·(−1)−1·1) = −(−3) = +3
C₃₁= +(1·2−(−1)·3) = +(2+3) = +5
C₃₂= −(2·2−(−1)·0) = −4
C₃₃= +(2·3−1·0) = +6
3
M⁻¹ = (1/33) × Cᵀ
Cᵀ= [[14, −3, 5], [2, 9, −4], [−3, 3, 6]](rows ↔ columns of C)
M⁻¹= (1/33)[[14, −3, 5], [2, 9, −4], [−3, 3, 6]]

Verify: check that M·M⁻¹ = I by multiplying row 1 of M by column 1 of M⁻¹: (2·14+1·2−1·(−3))/33 = (28+2+3)/33 = 33/33 = 1 ✓

Example 2 Transformation Matrix and Composition ★★☆ Challenging
Matrix A represents a reflection in the line y = x. Matrix B represents a rotation of 90° anticlockwise about the origin. (a) Write down A and B. (b) Find AB and BA. (c) Describe the transformation represented by AB. (d) Find the area scale factor of AB.
1
Part (a) — Write down A and B
A (reflect y=x)= [[0, 1], [1, 0]]
B (rotate 90° ACW)= [[0, −1], [1, 0]]
2
Part (b) — Compute AB and BA
AB= [[0,1],[1,0]]·[[0,−1],[1,0]] = [[0·0+1·1, 0·(−1)+1·0], [1·0+0·1, 1·(−1)+0·0]]
AB= [[1, 0], [0, −1]]B first, then A
BA= [[0,−1],[1,0]]·[[0,1],[1,0]] = [[−1,0],[0,1]]
3
Parts (c) and (d)
AB = [[1,0],[0,−1]]Reflection in the x-axis(x unchanged, y negated)
det(AB)= 1×(−1) − 0×0 = −1
Area scale factor= |det(AB)| = 1area unchanged

Also note det(AB) = det(A)·det(B) = (−1)(1) = −1 ✓. This confirms orientation is reversed (reflection).

Example 3 Invariant Points and Invariant Lines ★★★ A* Level
The matrix M = [[3, 1], [2, 2]] represents a transformation. (a) Find all invariant points of M. (b) Find all invariant lines through the origin of M. (c) Find the invariant line NOT through the origin of the form y = 2x + c.
1
Part (a) — Invariant points: M·v = v → (M−I)v = 0
M − I= [[3−1, 1], [2, 2−1]] = [[2, 1], [2, 1]]
System:2x + y = 0 and 2x + y = 0 (same equation)
y = −2xfor all x

Invariant points lie on the line y = −2x — every point on this line is fixed.

2
Part (b) — Invariant lines through origin: y = mx

Let point (t, mt) lie on y = mx. Apply M:

Image= [[3,1],[2,2]]·[t, mt]ᵀ = [3t + mt, 2t + 2mt]ᵀ = [t(3+m), 2t(1+m)]ᵀ
Requirey'/x' = m:   2t(1+m) / [t(3+m)] = m
2(1+m) = m(3+m)
2 + 2m = 3m + m²
m² + m − 2 = 0   →   (m+2)(m−1) = 0
m = −2 or m = 1Invariant lines: y = −2x and y = x

Note: y = −2x is the line of invariant points (from part a). y = x is an invariant line where points move along it but are not fixed.

3
Part (c) — Invariant line y = 2x + c, c ≠ 0

Wait — from part (b), the invariant lines through the origin have gradients m = −2 and m = 1. We need a line y = 2x + c. But m = 2 is not an eigenvalue. So lines y = 2x + c are not invariant for this matrix. Let us instead find all invariant lines y = x + c (gradient 1):

Point (t, t+c)on y = x + c
Image= [3t+(t+c), 2t+2(t+c)]ᵀ = [4t+c, 4t+2c]ᵀ
Require y' = x' + c:4t+2c = (4t+c) + c = 4t+2c ✓

This holds for any value of c — every line y = x + c is an invariant line! This makes sense because y = x (through origin) is an invariant line, and the transformation acts by translating along lines parallel to y = x.

Example 4 (AB)⁻¹ and Solving Matrix Equations ★★☆ Challenging
Given A = [[2, 3], [1, 2]] and B = [[1, −1], [0, 2]], find (a) det(A), A⁻¹, (b) (AB)⁻¹ using the reversal rule, (c) the matrix X such that AX = B.
1
Part (a)
det A= 2·2 − 3·1 = 4 − 3 = 1
A⁻¹= (1/1)[[2, −3], [−1, 2]] = [[2, −3], [−1, 2]]
2
Part (b) — (AB)⁻¹ = B⁻¹A⁻¹
det B= 1·2 − (−1)·0 = 2
B⁻¹= (1/2)[[2, 1], [0, 1]] = [[1, ½], [0, ½]]
(AB)⁻¹= B⁻¹A⁻¹ = [[1,½],[0,½]]·[[2,−3],[−1,2]]
= [[2−½, −3+1], [0−½, 0+1]] = [[3/2, −2], [−1/2, 1]]
3
Part (c) — Solve AX = B
AX = B→ X = A⁻¹Bpre-multiply both sides by A⁻¹
X= [[2,−3],[−1,2]]·[[1,−1],[0,2]]
= [[2·1+(−3)·0, 2·(−1)+(−3)·2], [(−1)·1+2·0, (−1)·(−1)+2·2]]
X= [[2, −8], [−1, 5]]

Interactive Transformation Laboratory

Enter a 2×2 matrix and watch how it transforms a unit square. Invariant lines are shown automatically.

🔬 2×2 Matrix Transformation Visualiser
—det M
—Area Scale Factor
—Singular?

Matrix Inverse Calculator

Practice Questions

Question 1 — 2×2 Operations and Inverse
Given A = [[3, −1], [5, 2]] and B = [[1, 2], [−1, 3]], find: (a) AB, (b) det(A) and A⁻¹, (c) (AB)⁻¹ using B⁻¹A⁻¹, (d) the matrix X such that XA = B.
AB: standard row × column multiplication. For A⁻¹: det A = 3·2−(−1)·5 = 11. For (d): XA = B → X = BA⁻¹ (post-multiply by A⁻¹ on the right).
✓ Solution
(a) AB= [[3·1+(−1)(−1), 3·2+(−1)·3], [5·1+2(−1), 5·2+2·3]] = [[4, 3], [3, 16]]
(b) det A= 6−(−5) = 11
A⁻¹= (1/11)[[2,1],[−5,3]]
(c) det B= 3+2 = 5, B⁻¹ = (1/5)[[3,−2],[1,1]]
(AB)⁻¹= B⁻¹A⁻¹ = (1/5)[[3,−2],[1,1]]·(1/11)[[2,1],[−5,3]] = (1/55)[[16,−3],[−3,4]]
(d) X = BA⁻¹= [[1,2],[−1,3]]·(1/11)[[2,1],[−5,3]] = (1/11)[[−8,7],[−17,8]]
Question 2 — Transformation Matrix
Find the 2×2 matrix that represents: (a) a rotation of 45° anticlockwise, (b) a reflection in the line y = −x, (c) the transformation which is reflection in the x-axis followed by rotation of 90° anticlockwise. For (c), also find the area scale factor and describe the combined transformation.
(a) Use rotation matrix with θ=45°: cos45°=1/√2, sin45°=1/√2. (b) Use column rule: (1,0)→(−1,0)? No — find image of (1,0) under reflection in y=−x. (c) "Rotation then Reflection" — but order: reflection first, then rotation → matrix = (rotation)·(reflection).
✓ Solution
(a)[[cos45°, −sin45°],[sin45°, cos45°]] = (1/√2)[[1,−1],[1,1]]
(b) y=−x:(1,0)→(0,−1), (0,1)→(−1,0) → M = [[0,−1],[−1,0]]
(c)Rotation R = [[0,−1],[1,0]], Reflection S = [[1,0],[0,−1]]
S then R:RS = [[0,−1],[1,0]]·[[1,0],[0,−1]] = [[0,1],[1,0]]
= reflection in y=xArea scale = |det| = |−1| = 1
Question 3 — Invariant Points and Lines
The matrix M = [[4, −1], [2, 1]] represents a transformation T. (a) Find the invariant points of T. (b) Find the equations of all invariant lines of T through the origin. (c) Show that any line of the form y = 2x + c is an invariant line of T.
(a) (M−I)v=0: M−I = [[3,−1],[2,0]]. (b) Let (t,mt) → image, require y'/x' = m, solve quadratic in m. (c) Let point (t, 2t+c) → apply M → check image satisfies y=2x+c.
✓ Solution
(a) M−I= [[3,−1],[2,0]]: gives 3x−y=0, 2x=0 → x=0, y=0
Invariant pointonly the origin (0,0)
(b) Image of (t,mt)= [4t−mt, 2t+mt] = [t(4−m), t(2+m)]
Require:t(2+m)/t(4−m) = m → 2+m = m(4−m) → m²−3m+2=0
(m−1)(m−2)=0m=1 or m=2 → y=x and y=2x
(c) Point (t,2t+c)→ image = [4t−(2t+c), 2t+(2t+c)] = [2t−c, 4t+c]
Check y=2x+c:4t+c = 2(2t−c)+c = 4t−2c+c = 4t−c ✗

⚠ Wait: 4t+c ≠ 4t−c unless c=0. So y=2x+c is NOT invariant for general c. But y=2x (c=0) is invariant (it's the m=2 line). Check c=0: image [2t,4t]: y=4t=2(2t) ✓. The question as stated has an error — y=2x (c=0) is invariant, not y=2x+c for c≠0.

Question 4 — A* 3×3 Matrix System
The matrix A = [[1, 2, 0], [3, 1, 1], [0, −1, 2]]. (a) Find det(A). (b) Find A⁻¹. (c) Hence solve the system: x+2y=5, 3x+y+z=10, −y+2z=3.
Expand det along row 1: a(ei−fh)−b(di−fg)+c(dh−eg). For the system, write it as A·[x,y,z]ᵀ = [5,10,3]ᵀ, then [x,y,z]ᵀ = A⁻¹·[5,10,3]ᵀ.
✓ Solution
(a) det A= 1(1·2−1·(−1)) − 2(3·2−1·0) + 0 = 1(3) − 2(6) = 3−12 = −9
Cofactors:C₁₁=3, C₁₂=−6, C₁₃=−3, C₂₁=−4, C₂₂=2, C₂₃=1, C₃₁=2, C₃₂=−1, C₃₃=−5
(b) A⁻¹= (1/−9)·Cᵀ = (−1/9)[[3,−4,2],[−6,2,−1],[−3,1,−5]]
(c) [x,y,z]ᵀ= A⁻¹·[5,10,3]ᵀ
x= (−1/9)(3·5+(−4)·10+2·3) = (−1/9)(15−40+6) = (−1/9)(−19) = 19/9
y= (−1/9)(−30+20−3) = (−1/9)(−13) = 13/9
z= (−1/9)(−15+10−15) = (−1/9)(−20) = 20/9

Formula Reference Sheet

Complete reference for Matrices — Cambridge 9231 P1, Section 1.4.

2×2 Determinant and Inverse
det[[a,b],[c,d]] = ad−bc
M⁻¹ = (1/det)[[d,−b],[−c,a]]
Singular: det = 0 (no inverse)
det(AB) = det(A)·det(B)
Inverse Properties
(AB)⁻¹ = B⁻¹A⁻¹reversal
(A⁻¹)⁻¹ = A
AX = B → X = A⁻¹B
XA = B → X = BA⁻¹
Standard Transformation Matrices
Rotation θ ACW: [[cosθ,−sinθ],[sinθ,cosθ]]
Reflect x-axis: [[1,0],[0,−1]]
Reflect y-axis: [[−1,0],[0,1]]
Reflect y=x: [[0,1],[1,0]]
Enlarge ×k: [[k,0],[0,k]]
Shear x: [[1,k],[0,1]]
Composition and Area
AB = "B then A"
Area scale = |det M|
det > 0: orientation preserved
det < 0: orientation reversed
Invariant Points
Mv = v → (M−I)v = 0
Origin always invariant
If det(M−I) = 0: line of inv. pts
Invariant Lines y = mx
Point (t, mt) → M → (x', y')
Require y'/x' = m
Solve quadratic in m
Roots give invariant line gradients
📋 Cambridge Exam Strategy — Matrices
  • Check the sign pattern for 3×3 cofactors before every calculation: + − + / − + − / + − +. One sign error cascades through the whole inverse.
  • For composition "B then A": the matrix is AB, not BA. Write the second transformation on the right.
  • For matrix equations: XA = B gives X = BA⁻¹ (post-multiply); AX = B gives X = A⁻¹B (pre-multiply). The position of A⁻¹ matters.
  • For invariant lines: always state whether each line through the origin has all points fixed or just the line as a whole invariant. These are different statements.
  • Verify 3×3 inverses by checking at least one entry of M·M⁻¹ equals the corresponding identity entry.
  • For solving 3×3 systems: write the system as A·x = b explicitly, then x = A⁻¹b. This is cleaner than elimination in exams.
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