Matrix Transformations – Grade 10 IGCSE Maths

Welcome to Matrix Transformations!

A matrix is a rectangular array of numbers. In IGCSE Maths, 2×2 matrices are used to represent geometric transformations — reflections, rotations, enlargements and shears — all via a single elegant multiplication. Understanding matrices unlocks a powerful unified language for all transformations.

Transformation matrix M maps point (x,y) to new point: M × [x; y] = [new x; new y]

Multiply a 2×2 matrix by a column vector to transform a point
Apply standard matrices for reflections, rotations, enlargements and shears
Combine two transformations using matrix multiplication (B×A order)
Calculate the determinant and inverse of a 2×2 matrix
Use the determinant to find the area scale factor of a transformation
Identify the geometric transformation from a given matrix

Matrix × Vector

Transforming points and shapes

Standard Matrices

Reflections, rotations, enlargements

Combined Transformations

B×A order — apply A then B

Determinant

ad − bc and area scale factor

Inverse Matrix

Reverse a transformation

Matrix Plotter

See transformations visually

1. Matrix Multiplication — Transforming a Point

A 2×2 matrix multiplied by a 2×1 column vector produces a new column vector — the image of the point after transformation.

[a b; c d] × [x; y] = [ax + by; cx + dy]

Example: Apply M = [2 1; 0 3] to point (1, 2):
New x = 2(1) + 1(2) = 4
New y = 0(1) + 3(2) = 6
Image point: (4, 6)

Transforming a shape: Apply the matrix to every vertex in turn.
Triangle with vertices (1,0), (0,1), (2,2): apply M to each vertex separately, then connect the image points.

2. The Identity Matrix

I = [1 0; 0 1] — maps every point to itself (no change)

The identity matrix is the matrix equivalent of multiplying by 1. Any matrix M satisfies: M × I = I × M = M.

3. Standard Transformation Matrices

Each geometric transformation has a unique 2×2 matrix. Verify by checking where (1,0) and (0,1) map to — these give the columns of the matrix.

Transformation	Matrix	Check: (1,0)→?
Reflection in x-axis	[1 0; 0 −1]	(1, 0) ✓
Reflection in y-axis	[−1 0; 0 1]	(−1, 0) ✓
Reflection in y = x	[0 1; 1 0]	(0, 1) ✓
Reflection in y = −x	[0 −1; −1 0]	(0, −1) ✓
Rotation 90° anticlockwise	[0 −1; 1 0]	(0, 1) ✓
Rotation 180°	[−1 0; 0 −1]	(−1, 0) ✓
Rotation 90° clockwise	[0 1; −1 0]	(0, −1) ✓
Enlargement scale factor k	[k 0; 0 k]	(k, 0) ✓
Stretch ×a parallel to x-axis	[a 0; 0 1]	(a, 0) ✓
Stretch ×b parallel to y-axis	[1 0; 0 b]	(1, 0) ✓
Shear parallel to x-axis, factor k	[1 k; 0 1]	(1, 0) ✓

Memory trick: the columns of the matrix tell you where (1,0) and (0,1) go. First column = image of (1,0). Second column = image of (0,1).

1. Combined Transformations — Order Matters!

When you apply transformation A first, then transformation B, the resultant single matrix is B × A — note the reverse order. Matrix multiplication is not commutative: BA ≠ AB in general.

Apply A first, then B: resultant matrix = B × A

Example: Reflect in x-axis (A), then rotate 90° anticlockwise (B).
A = [1 0; 0 −1] B = [0 −1; 1 0]

B × A = [0 −1; 1 0] × [1 0; 0 −1]
Row 1 of B × Col 1 of A: (0)(1)+(−1)(0) = 0
Row 1 of B × Col 2 of A: (0)(0)+(−1)(−1) = 1
Row 2 of B × Col 1 of A: (1)(1)+(0)(0) = 1
Row 2 of B × Col 2 of A: (1)(0)+(0)(−1) = 0
Result = [0 1; 1 0] — this is reflection in y = x ✓

2. Matrix Multiplication — Full Method

[a b; c d] × [e f; g h] = [ae+bg af+bh; ce+dg cf+dh]

Worked: [2 1; 3 0] × [1 4; 2 −1]
Top-left: 2(1)+1(2) = 4
Top-right: 2(4)+1(−1) = 7
Bottom-left: 3(1)+0(2) = 3
Bottom-right: 3(4)+0(−1) = 12
Result: [4 7; 3 12]

Each entry = (row of first matrix) · (column of second matrix). There are 4 dot products to compute for a 2×2 result.

3. Identifying the Single Equivalent Transformation

Once you have the combined matrix, compare it to the standard matrices table to identify what single geometric transformation it represents.

Strategy:
1. Check where (1,0) maps → first column of matrix
2. Check where (0,1) maps → second column of matrix
3. Ask: is it a reflection? rotation? enlargement?
4. For rotation: det = 1 (positive), angle = arctan(c/a)
5. For reflection: det = −1

1. Determinant of a 2×2 Matrix

det [a b; c d] = ad − bc

Example: det [3 2; 1 4] = (3)(4) − (2)(1) = 12 − 2 = 10
Example: det [0 −1; 1 0] = (0)(0) − (−1)(1) = 0 + 1 = 1 (rotation: det always 1)
Example: det [1 0; 0 −1] = (1)(−1) − (0)(0) = −1 (reflection: det always −1)

2. Area Scale Factor

Area of image = |det(M)| × Area of original

If triangle has area 6 cm² and matrix has det = 3, image area = 18 cm².
If det = −2, area scale factor = |−2| = 2 (negative just means orientation reverses).

Singular matrix: if det = 0, the transformation collapses the plane to a line (or point). The image is 1-dimensional — no inverse exists.

3. The Inverse Matrix

M⁻¹ = (1/det) × [d −b; −c a] — swap main diagonal, negate off-diagonal, divide by det

Example: Find the inverse of M = [3 1; 2 4]
det = (3)(4) − (1)(2) = 12 − 2 = 10
M⁻¹ = (1/10) × [4 −1; −2 3] = [0.4 −0.1; −0.2 0.3]

Verify: M × M⁻¹ = [1 0; 0 1] ✓

Application: if matrix M maps shape A to shape B, then M⁻¹ maps shape B back to shape A. Use inverse to "undo" a transformation.

Solving MX = B for X:
Multiply both sides on the LEFT by M⁻¹:
M⁻¹ × M × X = M⁻¹ × B
I × X = M⁻¹ × B
X = M⁻¹ × B

Example 1 — Apply Matrix to Triangle Vertices

Matrix M = [2 0; 1 3]. Triangle vertices: A(1,0), B(2,0), C(1,2).

A(1,0): new x = 2(1)+0(0)=2, new y = 1(1)+3(0)=1 → A'(2,1)

B(2,0): new x = 2(2)+0(0)=4, new y = 1(2)+3(0)=2 → B'(4,2)

C(1,2): new x = 2(1)+0(2)=2, new y = 1(1)+3(2)=7 → C'(2,7)

Image triangle: A'(2,1), B'(4,2), C'(2,7)

Example 2 — Identify Transformation from Matrix

Given M = [0 −1; 1 0]. What transformation does M represent?

Check (1,0): M×[1;0] = [0(1)+(−1)(0); 1(1)+0(0)] = [0; 1]. So (1,0)→(0,1).

Check (0,1): M×[0;1] = [0(0)+(−1)(1); 1(0)+0(1)] = [−1; 0]. So (0,1)→(−1,0).

det = (0)(0)−(−1)(1) = 1 (positive → rotation, not reflection).

Answer: Rotation of 90° anticlockwise about the origin.

Example 3 — Find Combined Transformation Matrix

Apply: first reflect in y-axis (A), then rotate 180° (B). Find B×A.

A = [−1 0; 0 1], B = [−1 0; 0 −1]

B×A = [−1 0; 0 −1] × [−1 0; 0 1]

Top-left: (−1)(−1)+(0)(0)=1. Top-right: (−1)(0)+(0)(1)=0.

Bottom-left: (0)(−1)+(−1)(0)=0. Bottom-right: (0)(0)+(−1)(1)=−1.

B×A = [1 0; 0 −1] — this is reflection in the x-axis.

Example 4 — Find the Inverse Matrix

M = [5 2; 3 1]. Find M⁻¹.

det(M) = (5)(1) − (2)(3) = 5 − 6 = −1

M⁻¹ = (1/−1) × [1 −2; −3 5] = [−1 2; 3 −5]

Verify: [5 2; 3 1]×[−1 2; 3 −5] = [5(−1)+2(3), 5(2)+2(−5); 3(−1)+1(3), 3(2)+1(−5)] = [1 0; 0 1] ✓

Example 5 — Calculate Area of Image

A triangle has area 8 cm². Transformation matrix M = [3 1; 2 2]. Find area of image.

det(M) = (3)(2) − (1)(2) = 6 − 2 = 4

Area scale factor = |det| = 4

Area of image = 4 × 8 = 32 cm²

Example 6 — Solve MX = Q for X

M = [2 1; 1 1], Q = [5; 3]. Find X such that MX = Q.

det(M) = (2)(1)−(1)(1) = 1, so M⁻¹ = [1 −1; −1 2]

X = M⁻¹ × Q = [1 −1; −1 2] × [5; 3]

X₁ = 1(5)+(−1)(3) = 2; X₂ = (−1)(5)+2(3) = 1

X = [2; 1], i.e., the point (2, 1). Check: M×[2;1]=[5;3] ✓

Common Mistakes to Avoid

Mistake 1 — Wrong Order for Combined Transformations

"Apply A then B" means the matrix is B × A, NOT A × B. Students often write A × B because they read left to right, but matrix multiplication reads right to left for "then". Always write the later transformation on the LEFT.

Check: [Rotate] × [Reflect] means reflect FIRST, then rotate.

Mistake 2 — Determinant Formula Error

det [a b; c d] = ad − bc, NOT ab − cd or ad + bc. The determinant uses the product of the main diagonal minus the product of the off-diagonal. The off-diagonal is b×c (top-right times bottom-left).

Common wrong answer: det [3 2; 1 4] = 3×2 − 1×4 = 2. Correct: det = 3×4 − 2×1 = 10.

Mistake 3 — Forgetting the 1/det Factor in the Inverse

Students swap the diagonal and negate correctly, but forget to multiply by 1/det. The inverse is (1/det)×[d −b; −c a]. If det = 3, every entry must be divided by 3.

Mistake 4 — Rotation Direction Confusion

The matrix [0 −1; 1 0] is anticlockwise 90°. Verify: it maps (1,0) → (0,1), which is a quarter-turn anticlockwise. The clockwise version [0 1; −1 0] maps (1,0) → (0,−1). Always check by applying to (1,0).

Mistake 5 — Area Scale Factor is |det|, Not det

The area scale factor is the absolute value |det(M)|. If det = −3, the area scale factor is 3 (not −3). A negative determinant means orientation reverses (e.g. clockwise becomes anticlockwise), but area is always positive.

Key Formulas — Matrix Transformations

Formula / Matrix	What it Does	Status
[a b; c d]×[x;y] = [ax+by; cx+dy]	Matrix × vector multiplication	Memorise
det = ad − bc	Determinant of 2×2 matrix	Memorise
M⁻¹ = (1/det)×[d −b; −c a]	Inverse of 2×2 matrix	Memorise
Area image = \|det\| × Area original	Area scale factor	Memorise
[1 0; 0 1]	Identity — no transformation	Reference
[1 0; 0 −1]	Reflection in x-axis	Memorise
[−1 0; 0 1]	Reflection in y-axis	Memorise
[0 1; 1 0]	Reflection in y = x	Memorise
[0 −1; −1 0]	Reflection in y = −x	Memorise
[0 −1; 1 0]	Rotation 90° anticlockwise	Memorise
[−1 0; 0 −1]	Rotation 180°	Memorise
[0 1; −1 0]	Rotation 90° clockwise	Memorise
[k 0; 0 k]	Enlargement scale factor k	Memorise
[a 0; 0 1]	Stretch ×a parallel to x-axis	Reference
[1 0; 0 b]	Stretch ×b parallel to y-axis	Reference
[1 k; 0 1]	Shear parallel to x-axis, factor k	Reference
Combined: apply A then B → B×A	Order rule for combined matrices	Memorise

Matrix Transformation Plotter

Enter 4 values of a 2×2 matrix. The canvas draws a unit square and triangle, applies the transformation, and shows the image. The determinant and area scale factor are calculated automatically.

a: b: c: d:

Matrix: [a b; c d] — shown below as [[a,b],[c,d]]

Enter matrix values and click Apply Transformation.

Exercise 1 — Matrix × Vector

Apply the given matrix to the given point. Enter the x-coordinate of the image.

1. M = [2 1; 0 3], point (1, 0). Enter image x-coord.

2. M = [2 1; 0 3], point (0, 1). Enter image x-coord.

3. M = [1 0; 0 −1] (reflect x-axis), point (3, 4). Enter image y-coord.

4. M = [−1 0; 0 1] (reflect y-axis), point (5, 2). Enter image x-coord.

5. M = [0 −1; 1 0] (rot 90° ACW), point (3, 0). Enter image y-coord.

6. M = [0 −1; 1 0], point (0, 4). Enter image x-coord.

7. M = [3 0; 0 3] (enlargement ×3), point (2, 5). Enter image x-coord.

8. M = [0 1; 1 0] (reflect y=x), point (3, 7). Enter image x-coord (= original y).

Exercise 2 — Transforming Triangle Vertices

Triangle has vertices A(2,0), B(0,2), C(2,2). Apply the given matrix. Enter the requested coordinate.

1. M = [1 0; 0 −1]. Give the y-coord of A'.

2. M = [1 0; 0 −1]. Give the y-coord of C'.

3. M = [0 1; 1 0] (y=x). Give the x-coord of B'.

4. M = [0 1; 1 0] (y=x). Give the x-coord of C'.

5. M = [2 0; 0 2] (enlargement ×2). Give x-coord of B'.

6. M = [2 0; 0 2]. Give y-coord of C'.

7. M = [0 −1; 1 0] (rot 90° ACW). A(2,0)→A'. Give x-coord of A'.

8. M = [−1 0; 0 −1] (rot 180°). C(2,2)→C'. Give x-coord of C'.

Exercise 3 — Identify the Transformation

For each matrix, apply it to (1,0) and (0,1) to identify the transformation. Enter the determinant.

1. M = [1 0; 0 −1]. What is det(M)? Enter as a number.

2. M = [−1 0; 0 1]. What is det(M)?

3. M = [0 −1; 1 0]. What is det(M)?

4. M = [3 0; 0 3]. What is det(M)?

5. M = [2 1; 1 2]. What is det(M)?

6. M = [4 2; 2 1]. What is det(M)? (Singular matrix — det = 0)

7. M = [3 1; 2 4]. det(M) = ?

8. M = [5 3; 2 2]. det(M) = ?

Exercise 4 — Finding the Inverse Matrix

Find M⁻¹ = (1/det)×[d −b; −c a]. Enter the top-left entry of the inverse matrix.

1. M = [2 1; 1 1]. det = 1. Top-left of M⁻¹ = ?

2. M = [3 1; 2 1]. det = 1. Top-left of M⁻¹ = ?

3. M = [4 1; 2 1]. det = 2. Top-left of M⁻¹ = (1/det)×d = 0.5×1 = ?

4. M = [5 2; 3 1]. det = −1. Top-left of M⁻¹ = (1/det)×d = −1×1 = ?

5. M = [2 3; 1 2]. det = 1. Top-left of M⁻¹ = d = ?

6. M = [3 2; 1 1]. det = 1. Top-left of M⁻¹ = d = ?

7. M = [1 0; 0 −1] (reflect x-axis). Top-left of M⁻¹ = ? (Same matrix, since reflections are self-inverse.)

8. M = [0 −1; 1 0] (rot 90° ACW). det = 1. M⁻¹ top-left = d = 0. Top-right = −b = 1. Enter top-right.

Exercise 5 — Area Scale Factor

Use area of image = |det(M)| × original area.

1. M = [2 0; 0 2], original area = 5. Image area = ?

2. M = [3 0; 0 3], original area = 4. Image area = ?

3. M = [1 0; 0 −1], original area = 10. Image area = ?

4. M = [2 1; 1 2], det = 3, original area = 6. Image area = ?

5. M = [3 1; 2 4], det = 10, original area = 5. Image area = ?

6. M = [5 2; 3 1], det = −1, original area = 8. Image area = ?

7. M = [4 1; 2 3], det = 10, original area = 3. Image area = ?

8. M = [0 −1; 1 0] (rot 90° ACW), original area = 15. Image area = ? (det = 1)

Practice — 25 Questions

Mixed practice on all matrix transformation skills.

1. M=[2 0;0 2], point (3,4). Image x-coord?

2. M=[1 0;0 −1], point (2,5). Image y-coord?

3. M=[−1 0;0 1], point (4,3). Image x-coord?

4. M=[0 1;1 0], point (7,2). Image x-coord?

5. M=[0 −1;1 0], point (5,0). Image y-coord?

6. det [3 2;1 4] = ?

7. det [5 3;2 2] = ?

8. det [0 −1;1 0] = ?

9. M=[2 1;1 1]. Top-left of M⁻¹ (det=1)?

10. Area scale factor if det(M) = 5?

11. M=[3 0;0 3], original area=7. Image area?

12. det [4 2;2 1] = ? (singular matrix)

13. M=[2 1;3 2], det=1. Top-left of M⁻¹ = d = ?

14. M=[1 0;0 −1] applied to (3,−2). Image y-coord?

15. Combine: A=[1 0;0 −1], B=[0 −1;1 0]. Compute B×A top-left entry.

16. B×A top-right for A=[1 0;0 −1], B=[0 −1;1 0].

17. M=[2 3;1 2]. det=1. Bottom-right of M⁻¹ = a = ?

18. M=[3 1;2 4], original area=4. Image area (det=10)?

19. M=[0 −1;1 0], point (2,3). Image x-coord?

20. M=[0 −1;1 0], point (2,3). Image y-coord?

21. M=[−1 0;0 −1] (rot 180°), point (4,−3). Image y-coord?

22. M=[5 2;3 1], det=−1. Top-left of M⁻¹ = (1/det)×d = −1?

23. det [6 4;3 2] = ? (singular)

24. M=[2 0;0 3], point (4,2). Image y-coord?

25. M=[1 2;0 1] (shear), point (3,4). Image x-coord?

Challenge — 12 Multi-Step Questions

IGCSE-style multi-step matrix problems. Show working on paper where needed.

1. M=[2 1;1 1]. Find det(M). Enter it.

2. Using M=[2 1;1 1] above, find top-right of M⁻¹ = (1/det)×(−b) = −1.

3. A=[0 −1;1 0], B=[1 0;0 −1]. Compute B×A. What is top-left entry of B×A?

4. Same B×A. What is bottom-right entry?

5. Triangle area = 12 cm². M=[3 1;2 4], det=10. Image area = ?

6. M=[4 1;3 1], det=1. Bottom-left of M⁻¹ = (1/det)×(−c) = −3.

7. MX=[5;3], M=[2 1;1 1], M⁻¹=[1 −1;−1 2]. X₁=M⁻¹×[5;3] top entry = 1×5+(−1)×3 = ?

8. Same, X₂ = (−1)×5+2×3 = ?

9. M=[3 2;1 1]. det(M)=?

10. Using det=1 for M=[3 2;1 1], top-left of M⁻¹=d=1. Bottom-right of M⁻¹ = a = ?

11. Reflect in x-axis (A), then reflect in y=x (B). B×A top-left entry: B=[0 1;1 0], A=[1 0;0 −1]. Compute.

12. Same B×A, bottom-right entry.

Exam Style — 5 Structured Questions

Multi-part IGCSE examination questions. Enter numerical answers where prompted.

Q1(a). M = [2 3; 1 2]. Calculate det(M).

Q1(b). Using det = 1, find the top-right entry of M⁻¹ = (1/1)×[d −b; −c a]. Top-right = −b = −3.

Q2. Matrix A = [1 0; 0 −1] (reflection x-axis), B = [0 −1; 1 0] (rotation 90° ACW). "Apply A then B" means compute B×A. Enter the bottom-left entry of B×A.

Q3. Triangle with area 8 cm² is transformed by M = [3 1; 2 2]. det(M) = 3×2 − 1×2 = 4. Area of image triangle = ?

Q4. MX = [7; 4] where M = [3 1; 2 1], det = 1, M⁻¹ = [1 −1; −2 3]. X = M⁻¹×[7;4]. Enter X₁ (first component of X).

Matrix Transformations GCSE Level