IB Maths AI HLMatrix TransformationsPaper 1 & 2Composite ST~9 min read
Matrices of Composite Transformations
Two transformations T then S applied to a point v give S(Tv) = (ST)v. So the single matrix for the composite is the product ST — second matrix on the left, first matrix on the right. Order matters: in general ST ≠ TS. Same transformation applied n times is Tn.
📘 What you need to know
Composite rule: T first, then S → matrix = ST. Pre-multiply by the next matrix, never post-multiply.
Three or more: T1 first, then T2, then T3 → matrix = T3·T2·T1.
Repeated: T applied n times = Tn. Use GDC for big n.
Order matters: ST ≠ TS in general — a rotation followed by a reflection is not the same as a reflection followed by the rotation.
Identity from periodicity: any transformation that “undoes itself after n repeats” gives Tn = I, where I = (1001).
· Reflection: S2 = I (do it twice → back to start)
· Rotation by θ such that nθ = 360°: Rn = I (e.g. 30° rotation → R12 = I)
Reverse direction: image P′ back to object P uses (ST)−1 = T−1S−1 — inverses swap order.
Inverse of a power: (Tn)−1 means “invert the whole thing”, not T−n. Compute Tn first, then invert.
Order matters — ST means “T first”
Reading right-to-left, the matrix closest to the vector acts first. So in (ST)v = S(Tv), T grabs v first, then S transforms the result. Swap the order and the answer changes — check with a quick example: rotation R by 90° anticlockwise, reflection F in the x-axis.
Composite matrix rule
v ⟶T Tv ⟶S S(Tv) = (ST)v
order of matrices reads left-to-right (S, T), order of action reads right-to-left (T first, S second)
Top: R first then F lands the triangle in Q3 (final A). Bottom: F first then R lands it in Q1 (final B). Same two transformations, different order, different image — the matrix product reflects this: FR ≠ RF.
Powers Tn and the identity trick
If you apply the same T n times, the combined matrix is Tn = T · T · … · T (n copies). Two useful periodicity facts:
Periodicity identities
S2 = I (any reflection S applied twice)
Rn = I when nθ = 360° (rotation R by angle θ about origin)
e.g. θ = 30° ⇒ n = 12, so R12 = I; θ = 90° ⇒ R4 = I
Why this helps: T50 looks scary, but if T4 = I then T50 = T4·12 + 2 = (T4)12 · T2 = I · T2 = T2. Reduce the exponent mod the period first.
🧠Recipe — composite transformations
List the steps in order — T1 first, T2 second, …, Tn last.
Get each matrix from the formula booklet (or derive from images of (1, 0) and (0, 1)).
Multiply in reverse order: composite C = Tn · … · T2 · T1.
Apply to the object: C · P gives the position matrix of the image.
Repeated application: use Tk (GDC). Reduce the exponent mod the period if Tperiod = I.
Reverse (image back to object): use C−1 — the GDC inverts in one step.
Worked examples
WE 1
Composite of two given matrices
T = (2001), S = (1101). Find the single matrix C for “T then S”.
T first, S second → C = STC = (1101)(2001)row × column(1,1): 1·2 + 1·0 = 2(1,2): 1·0 + 1·1 = 1(2,1): 0·2 + 1·0 = 0(2,2): 0·0 + 1·1 = 1C = (2101)
WE 2
Rotation then enlargement (formula booklet)
R = rotation 90° anticlockwise about origin. E = enlargement scale factor 0.25, centre origin. Find C = matrix for “R then E”.
write each matrixR = (cos 90°−sin 90°sin 90°cos 90°) = (0−110)E = (0.25000.25)R first, E second → C = ERC = (0.25000.25)(0−110)C = (0−0.250.250)scalar 0.25 multiplies every entry of R — quick shortcut here.
WE 3
Show ST ≠ TS
S = (100−1) (reflection in x-axis), T = (0−110) (rotation 90° CCW). Compute ST and TS, and confirm they differ.
STST = (100−1)(0−110) = (0−1−10)TSTS = (0−110)(100−1) = (0110)ST ≠ TS ✓ST is reflection in y = −x; TS is reflection in y = x. Same pieces, different result.
WE 4
Apply T four times — power of a matrix
C = (0−0.250.250). A square has position matrix P0 = (0025625602562560). Find P4 = C4P0.
C2 firstC2 = (0−0.250.250)2 = (−1/1600−1/16)C4 = (C2)2C4 = (1/256001/256)P4 = C4 · P0: scale each entry by 1/256P4 = (00110110)GDC: type C^4 · P0 in one calculation — avoids step-by-step slips.
WE 5
Reduce a large power using periodicity
R is rotation 60° anticlockwise about origin. Find R50.
Read ST right-to-left: T acts on the object first, then S.
Use the GDC for ST, Tn, T−1 — faster and fewer slips than by hand.
Check periodicity before computing large powers: reflection2 = I; rotation by θ has period 360°/θ.
Inverses swap order: (ST)−1 = T−1S−1. Same logic — undo S first, then undo T.
Verify direction: apply ST to one object point and check the image matches the expected geometric result.
âš Common mistakes
Wrong order: writing TS when the question says “T first, then S” — the matrix product is ST, with S on the left.
Assuming ST = TS — matrix multiplication is not commutative in general. Always check the order.
Treating (Tn)−1 as T−n — the question asks “invert the result of applying T n times”. Compute Tn first, then invert.
Forgetting to invert the order for inverses: (ST)−1 = T−1S−1, not S−1T−1.
Computing T50 step-by-step when periodicity gives a shortcut — always reduce the exponent mod the period when Tk = I.
Next up — Determinant of a Transformation Matrix. The area scale factor of any transformation is |det T| — so the image of a shape has area |det T| × (original area). A negative det T means the orientation flips (a reflection happened).
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