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Matrices and Linear Transformations

Year 1 / ASYear 2 / A-Level AQA, Edexcel, OCR, WJEC/Eduqas, CCEA

2x2 matrix multiplication, determinants, inverses, systems of equations and linear transformations with invariant points and lines - Edexcel 9FM0 Core Pure 1 revision with worked examples and mark-scheme logic.

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šŸŽÆ Learning Objectives

šŸ“‹ Key Definitions and Core Concepts

Matrix: A rectangular array of numbers; a 2x2 matrix has the form [[a, b], [c, d]] with four entries.
Determinant: For M = [[a, b], [c, d]], det M = ad āˆ’ bc. For a transformation, |det M| is the area scale factor; det M < 0 means orientation is reversed.
Singular matrix: A matrix with det M = 0: it has no inverse and its transformation collapses the plane onto a line or point.
Inverse matrix: M⁻¹ = (1/(ad āˆ’ bc))[[d, āˆ’b], [āˆ’c, a]], defined only when det M ≠ 0, satisfying MM⁻¹ = M⁻¹M = I.
Identity matrix: I = [[1, 0], [0, 1]]: AI = IA = A for every 2x2 matrix A.
Linear transformation: A map of the plane sending each position vector x to Mx; straight lines map to straight lines and the origin maps to itself.
Invariant point: A point whose image is itself: Mx = x, equivalently (M āˆ’ I)x = 0.
Invariant line: A line whose image under the transformation is the same line, though points on it may move along it; a line through the origin of gradient m satisfies M(1, m)įµ€ = k(1, m)įµ€ for some scalar k.

šŸ“Œ Key Points

AB ≠ BA in general - matrix multiplication is not commutative.
det(AB) = det A Ɨ det B; |det M| is the area scale factor of the transformation.
A 2x2 matrix is invertible if and only if ad āˆ’ bc ≠ 0.
Matrix of T followed by S is ST - the right-hand matrix acts first.
Rotations have det 1, reflections det āˆ’1, enlargement factor k has det k².
For a reflection, the mirror line is a line of invariant points; the perpendicular through O is an invariant line.

Matrix Arithmetic and Determinants

Products. For 2x2 matrices A = [[a, b], [c, d]] and B = [[e, f], [g, h]], the product AB has (i, j) entry given by multiplying row i of A into column j of B: AB = [[ae + bg, af + bh], [ce + dg, cf + dh]]. Multiplication is associative (A(BC) = (AB)C) and distributes over addition, but is not commutative: in general AB ≠ BA. The identity I = [[1, 0], [0, 1]] satisfies AI = IA = A. When a question gives a matrix equation such as A² = A (idempotent) or asks you to find matrix X satisfying AX = B, rearrange by multiplying by inverses on the correct side - left-multiplication and right-multiplication are different operations.

Worked example. A = [[2, 1], [5, 3]], B = [[0, 2], [3, 1]]: AB = [[2Ɨ0 + 1Ɨ3, 2Ɨ2 + 1Ɨ1], [5Ɨ0 + 3Ɨ3, 5Ɨ2 + 3Ɨ1]] = [[3, 5], [9, 13]]. Note BA = [[10, 6], [11, 6]] - different, as expected.

Determinants. det M = ad āˆ’ bc, also written det(M) or |M|. Two properties carry most of the mark weight: det(AB) = det A Ɨ det B, and det(Aįµ€) = det A. In the example above, det A = 2Ɨ3 āˆ’ 1Ɨ5 = 1 and det B = 0Ɨ1 āˆ’ 2Ɨ3 = āˆ’6, so det(AB) = āˆ’6 - and indeed 3Ɨ13 āˆ’ 5Ɨ9 = 39 āˆ’ 45 = āˆ’6 āœ“. Exam questions frequently ask you to verify this product rule.

Geometric meaning. For a transformation with matrix M, the image of a shape has area multiplied by |det M|, and det M < 0 signals that orientation is reversed (a flip). If det M = 0 the matrix is singular: the transformation squashes the whole plane onto a line (or a point), so distinct points share images and no inverse can exist. Singular matrices are the gateway to the system-of-equations cases in the next section.

Common trap

Entries go row-then-column: (row of A) Ɨ (column of B). Writing the transpose of the answer - columns of A times rows of B - is the single most common slip in matrix multiplication questions.

Inverses and Systems of Equations

The inverse formula. If det M = ad āˆ’ bc ≠ 0, then M⁻¹ = (1/(ad āˆ’ bc))[[d, āˆ’b], [āˆ’c, a]]: swap the leading diagonal, negate the other diagonal, divide by the determinant. Always check MM⁻¹ = I if asked to verify. The reversal rule (AB)⁻¹ = B⁻¹A⁻¹ holds whenever both factors are invertible - order flips, exactly as for transposes. A matrix is invertible if and only if it is non-singular.

Solving linear systems. The system 2x + 3y = 13, x āˆ’ 2y = āˆ’4 is Mx = b with M = [[2, 3], [1, āˆ’2]] and b = (13, āˆ’4)įµ€. det M = āˆ’4 āˆ’ 3 = āˆ’7, so M⁻¹ = (1/7)[[2, 3], [1, āˆ’2]] and x = M⁻¹b = (1/7)[[2, 3], [1, āˆ’2]](13, āˆ’4)įµ€ = (1/7)(26 āˆ’ 12, 13 + 8)įµ€ = (2, 3)įµ€. Substituting back: 2Ɨ2 + 3Ɨ3 = 13 āœ“. State the matrix equation, the determinant, the inverse and the solution - each step typically carries a mark.

Singular cases. When det M = 0 there is no unique solution, and the two possibilities are distinguished by consistency. The system x + 2y = 3, 2x + 4y = 7 has det [[1, 2], [2, 4]] = 0; doubling the first equation predicts 2x + 4y = 6, which contradicts 7, so the lines are parallel and distinct: no solution (inconsistent). But x + 2y = 3, 2x + 4y = 6 has the second equation a multiple of the first: the two equations describe the same line, so there are infinitely many solutions (a line of solutions). For 3x3 systems - which appear in OCR MEI Further Pure - the same trichotomy holds: unique (det ≠ 0), no common point, or a line/plane of common solutions.

Exam technique

Geometric interpretation marks are free if you learn the vocabulary: non-singular → unique intersection point; singular and consistent → lines coincide (infinitely many solutions); singular and inconsistent → parallel distinct lines (no solution). Quote the determinant first in every case.

Linear Transformations, Invariant Points and Invariant Lines

Matrices as maps. A 2x2 matrix M acts on position vectors: the point with position vector x maps to Mx. The unit square maps to a parallelogram whose sides are the columns of M - reading off the images of (1, 0) and (0, 1) is the fastest way to identify or construct a transformation. The standard matrices must be memorised:

TransformationMatrix
Rotation Īø anticlockwise about O[[cos Īø, āˆ’sin Īø], [sin Īø, cos Īø]]
Reflection in line through O at angle α to x-axis[[cos 2α, sin 2α], [sin 2α, āˆ’cos 2α]]
Enlargement, centre O, factor k[[k, 0], [0, k]]
Stretch parallel to x-axis, factor k[[k, 0], [0, 1]]
Stretch parallel to y-axis, factor k[[1, 0], [0, k]]

Check the reflection formula with y = x (α = 45°): [[cos 90°, sin 90°], [sin 90°, āˆ’cos 90°]] = [[0, 1], [1, 0]] āœ“. A negative enlargement factor k combines an enlargement with a 180° rotation, and has det k² > 0. Determinants classify the map: rotation det = 1 (area preserved), reflection det = āˆ’1 (area preserved, orientation reversed), enlargement det = k².

Composites. The matrix of transformation T followed by S is ST - the matrix acting first sits on the right. With RĪø denoting rotation by Īø, note that RĪøRφ = R(Īø+φ) but a reflection followed by a rotation is generally not a rotation: order matters both for the type of map and for its parameters.

Invariant points and lines. A point is invariant when Mx = x, i.e. (M āˆ’ I)x = 0. An invariant line maps to itself as a set - points may slide along it. For a line through the origin with gradient m, require M(1, m)įµ€ = k(1, m)įµ€. Worked example: M = [[3, āˆ’1], [āˆ’1, 3]]. Points on y = x: M(t, t)įµ€ = (2t, 2t)įµ€ - still on y = x, so y = x is an invariant line with scale factor 2 (only the origin is fixed). For general m: M(1, m)įµ€ = (3 āˆ’ m, āˆ’1 + 3m)įµ€ = k(1, m)įµ€, so k = 3 āˆ’ m and āˆ’1 + 3m = m(3 āˆ’ m), giving m² = 1: the invariant lines are y = x (scale 2) and y = āˆ’x (scale 4).

Structure to quote for common maps: for a reflection, the mirror line consists entirely of invariant points, and the line through O perpendicular to it is invariant with no fixed points (except O); for an enlargement with k ≠ 1, only O is an invariant point but every line through O is an invariant line.

Mark-scheme habit

Invariant line questions: take a general point (t, mt) or vector (1, m)įµ€, apply M, demand the image lie on the same line, solve for m, then state the scale factor k by comparing M(1, m)įµ€ with (1, m)įµ€. Distinguish 'invariant line' (set-wise) from 'line of invariant points' (point-wise) - examiners ask for exactly one of these.

šŸ’” Worked Example

Exam-style: The transformation T is represented by the matrix M = [[3, āˆ’1], [āˆ’1, 3]]. (a) Find det M and M⁻¹. (3 marks) (b) Show that y = x is an invariant line of T and state the scale factor of T on this line. (3 marks) (c) Find the equation of the other invariant line of T. (3 marks)
Mark-scheme logic: (a) det M = 3Ɨ3 āˆ’ (āˆ’1)(āˆ’1) = 9 āˆ’ 1 = 8. M⁻¹ = (1/8)[[3, 1], [1, 3]]. (b) Take a general point (t, t) on y = x: M(t, t)įµ€ = (3t āˆ’ t, āˆ’t + 3t)įµ€ = (2t, 2t)įµ€, which also lies on y = x. So y = x maps to itself - it is an invariant line - and each point on it is scaled by factor 2 about O (so only the origin is an invariant point on it). (c) Require M(1, m)įµ€ = k(1, m)įµ€: k = 3 āˆ’ m and āˆ’1 + 3m = km = m(3 āˆ’ m). Hence āˆ’1 + 3m = 3m āˆ’ m², so m² = 1 and m = ±1. m = 1 gives y = x from (b); the other invariant line is y = āˆ’x. (On it, M(1, āˆ’1)įµ€ = (4, āˆ’4)įµ€ = 4(1, āˆ’1)įµ€, scale factor 4.)

ā“ Practice Questions & Mark Schemes

A = [[2, 1], [5, 3]] and B = [[0, 2], [3, 1]]. Find AB and det B. (4 marks)
Answer: AB = [[2Ɨ0 + 1Ɨ3, 2Ɨ2 + 1Ɨ1], [5Ɨ0 + 3Ɨ3, 5Ɨ2 + 3Ɨ1]] = [[3, 5], [9, 13]]. det B = 0Ɨ1 āˆ’ 2Ɨ3 = āˆ’6. (Consistency check: det A = 1, so det(AB) = 1 Ɨ (āˆ’6) = āˆ’6, and 3Ɨ13 āˆ’ 5Ɨ9 = āˆ’6 āœ“.)
Find the inverse of M = [[3, āˆ’4], [1, 2]]. (3 marks)
Answer: det M = 3Ɨ2 āˆ’ (āˆ’4)Ɨ1 = 6 + 4 = 10 ≠ 0, so M is non-singular. M⁻¹ = (1/10)[[2, 4], [āˆ’1, 3]]. (Check: M Ɨ (1/10)[[2, 4], [āˆ’1, 3]] = (1/10)[[6 āˆ’ 4, 12 āˆ’ 12], [2 + 2, 4 + 6]] = I āœ“.)
Use a matrix method to solve the system 4x āˆ’ 3y = 6, 2x + y = 7. (4 marks)
Answer: Write Mx = b with M = [[4, āˆ’3], [2, 1]], b = (6, 7)įµ€. det M = 4 + 6 = 10, so M⁻¹ = (1/10)[[1, 3], [āˆ’2, 4]]. Then x = M⁻¹b = (1/10)(6 + 21, āˆ’12 + 28)įµ€ = (27/10, 8/5)įµ€. Check: 4(2.7) āˆ’ 3(1.6) = 10.8 āˆ’ 4.8 = 6 āœ“ and 2(2.7) + 1.6 = 7 āœ“.
Write down the matrix representing a rotation of 60° anticlockwise about the origin, and find the image of the point (1, 0). (3 marks)
Answer: [[cos 60°, āˆ’sin 60°], [sin 60°, cos 60°]] = [[1/2, āˆ’āˆš3/2], [√3/2, 1/2]]. The image of (1, 0) is the first column: (1/2, √3/2)įµ€. (det = 1/4 + 3/4 = 1, as expected for a rotation.)
Find all invariant points of the transformation with matrix M = [[2, āˆ’1], [āˆ’1, 2]]. (4 marks)
Answer: Invariant points satisfy Mx = x, i.e. (M āˆ’ I)x = 0 with M āˆ’ I = [[1, āˆ’1], [āˆ’1, 1]]. Both rows give the same equation x āˆ’ y = 0, so the solution set is the whole line y = x. Check: M(1, 1)įµ€ = (1, 1)įµ€ āœ“. Every point on y = x is fixed, and there are no other invariant points.

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