Matrices and Linear Transformations
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.
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.
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 ā 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 = [[2Ć0 + 1Ć3, 2Ć2 + 1Ć1], [5Ć0 + 3Ć3, 5Ć2 + 3Ć1]] = [[3, 5], [9, 13]]. Note BA = [[10, 6], [11, 6]] - different, as expected.
Determinants. det M = ad ā bc, also written det(M) or |M|. Two properties carry most of the mark weight: det(AB) = det A Ć det B, and det(Aįµ) = det A. In the example above, det A = 2Ć3 ā 1Ć5 = 1 and det B = 0Ć1 ā 2Ć3 = ā6, so det(AB) = ā6 - and indeed 3Ć13 ā 5Ć9 = 39 ā 45 = ā6 ā. 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.
Entries go row-then-column: (row of A) Ć (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.
The inverse formula. If det M = ad ā bc ā 0, then Mā»Ā¹ = (1/(ad ā bc))[[d, āb], [āc, a]]: swap the leading diagonal, negate the other diagonal, divide by the determinant. Always check MMā»Ā¹ = I if asked to verify. The reversal rule (AB)ā»Ā¹ = Bā»Ā¹Aā»Ā¹ 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 ā 2y = ā4 is Mx = b with M = [[2, 3], [1, ā2]] and b = (13, ā4)įµ. det M = ā4 ā 3 = ā7, so Mā»Ā¹ = (1/7)[[2, 3], [1, ā2]] and x = Mā»Ā¹b = (1/7)[[2, 3], [1, ā2]](13, ā4)įµ = (1/7)(26 ā 12, 13 + 8)įµ = (2, 3)įµ. Substituting back: 2Ć2 + 3Ć3 = 13 ā. 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 ā 0), no common point, or a line/plane of common solutions.
Geometric interpretation marks are free if you learn the vocabulary: non-singular ā unique intersection point; singular and consistent ā lines coincide (infinitely many solutions); singular and inconsistent ā parallel distinct lines (no solution). Quote the determinant first in every case.
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:
| Transformation | Matrix |
|---|---|
| Rotation Īø anticlockwise about O | [[cos Īø, āsin Īø], [sin Īø, cos Īø]] |
| Reflection in line through O at angle α to x-axis | [[cos 2α, sin 2α], [sin 2α, ācos 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°, ācos 90°]] = [[0, 1], [1, 0]] ā. 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 = ā1 (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 RĪøRĻ = R(Īø+Ļ) 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 ā 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, ā1], [ā1, 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 ā m, ā1 + 3m)įµ = k(1, m)įµ, so k = 3 ā m and ā1 + 3m = m(3 ā m), giving m² = 1: the invariant lines are y = x (scale 2) and y = āx (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 ā 1, only O is an invariant point but every line through O is an invariant line.
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.