IB Maths AI HL Matrix Transformations Paper 1 & 2 Image coordinates ~8 min read

Transformation by a Matrix

A transformation matrix maps every point of an object to a new point — its image — using the rule T(x, y) + (e, f) = (x′, y′). Each object point is written as a column vector, multiplied by the 2×2 matrix T, then shifted by the translation vector. To work backwards from image to object, subtract the translation and pre-multiply by T⁻¹.

📘 What you need to know

General form: T(xy) + (ef) = (x′y′) — matrix multiplication first, translation second.
The matrix T is a 2×2 array (abcd). It encodes rotations, reflections, stretches, enlargements — the translation vector (e, f) is separate.
Matrix × column vector: (abcd)(xy) = (ax + bycx + dy). Row of T × column of vector.
Position matrix P: stack several object points as columns of one matrix. Then T · P gives every image at once — called P′. Each column of P′ is one image coordinate.
Reversing the transformation: subtract the translation, then pre-multiply by T⁻¹. So (xy) = T⁻¹(x′ − ey′ − f).
GDC does the heavy lifting: enter T and P as matrices, compute T·P (and inverse) directly — especially useful when several points or non-integer entries are involved.
Order matters: T · P is not P · T. Always pre-multiply the column-vector / position matrix by T.

Applying a transformation matrix

Given an object point (x, y), the matrix transformation produces an image point (x′, y′) in three clean steps: write the object as a column vector, multiply by T (row-of-T dotted with the vector, twice), then add the translation. For several points at once, build the position matrix P with one point per column — this lets the GDC compute every image in a single matrix multiplication.

The blue triangle ABC is mapped to the teal triangle A′B′C′ by T = (2001) followed by translation (1, 0) — a horizontal stretch by factor 2, then a shift right by 1.

Matrix transformation formula (abcd)(xy) + (ef) = (x′y′) reverse direction: (xy) = T⁻¹(x′ − ey′ − f)

Working backwards — image to object

If you know an image point (x′, y′) and want the original (x, y), undo the transformation in reverse order. Translation came last, so subtract it first — that gives T(xy) = (x′ − ey′ − f). Then pre-multiply both sides by T⁻¹. The 2×2 inverse formula is in the formula booklet: for T = (abcd), T⁻¹ = 1det T(d−b−ca) where det T = ad − bc.

GDC tip: enter T as a 2×2 matrix and your column vector (or full position matrix) as another. The calculator handles T · P, T⁻¹, even Tⁿ directly — faster and less error-prone than doing the multiplications by hand. Always double-check the matrix entries before computing.

🧭 Recipe — finding the image of a point or shape

Write the object as a column vector (one point) or position matrix P (several points stacked as columns).
Multiply by T: compute T · P. Each entry is (row of T) · (column of P).
Add the translation vector (e, f) to every column — the result is the image position matrix P′.
Read off the image coordinates from the columns of P′.
To reverse (image → object): subtract the translation, then pre-multiply by T⁻¹.

Worked examples

WE 1

Apply a matrix to a single point (no translation)

The transformation matrix T = (2013) is applied to the point P(3, 2). Find the coordinates of the image P′.

write P as a column vector P = (32) compute T · P row by row top row: 2(3) + 0(2) = 6 bottom row: 1(3) + 3(2) = 9 P′ = (69) P′ = (6, 9) no translation here — just one matrix multiplication.

WE 2

Apply a matrix and a translation vector

The point Q(1, 4) is transformed by T = (1−121) with translation vector (3−2). Find the image Q′.

multiply T by Q first top row: 1(1) + (−1)(4) = −3 bottom: 2(1) + 1(4) = 6 T·Q = (−36) add the translation (−3 + 3, 6 + (−2)) = (0, 4) Q′ = (0, 4) translation always after the matrix multiplication — never before.

WE 3

Transform a triangle using the position matrix

Triangle has vertices A(1, 2), B(3, 1), C(2, 4). Apply T = (2001) with translation (10). Find the image coordinates.

build position matrix P (each column = one vertex) P = (132214) compute T·P (top row × P, bottom row × P) top: 2(1), 2(3), 2(2) → 2, 6, 4 bot: 1(2), 1(1), 1(4) → 2, 1, 4 T·P = (264214) add (1, 0) to every column P′ = (375214) A′(3, 2), B′(7, 1), C′(5, 4) one matrix multiplication produces all three images — GDC handles this in seconds.

WE 4

Find the original point given the image

Under the transformation T = (1235) with no translation, a point P has image (7, 18). Find the coordinates of P.

set up the equation T·P = (718) find T⁻¹: det T = 1(5) − 2(3) = −1 T⁻¹ = (1/−1)(5−2−31) = (−523−1) pre-multiply image by T⁻¹ top: −5(7) + 2(18) = −35 + 36 = 1 bot: 3(7) + (−1)(18) = 21 − 18 = 3 P = (1, 3) check: T(1,3) = (1+6, 3+15) = (7, 18) ✓

WE 5

Transform a quadrilateral with four vertices

A quadrilateral Q has vertices A(2, 5), B(5, 9), C(11, 9), D(8, 5). Find the coordinates of the image of Q under the transformation T = (3−1−12).

position matrix P (columns = A, B, C, D) P = (251185995) compute T·P, column by column col 1: 3(2)−5 = 1, −2+2(5) = 8 col 2: 3(5)−9 = 6, −5+2(9) = 13 col 3: 3(11)−9 = 24, −11+2(9) = 7 col 4: 3(8)−5 = 19, −8+2(5) = 2 P′ = (16241981372) A′(1, 8), B′(6, 13), C′(24, 7), D′(19, 2) GDC: enter T and P, compute T·P in one step.

WE 6

Reverse a transformation with a translation

The transformation T = (2111) with translation vector (1−3) sends a point P to the image P′(9, 5). Find the coordinates of P.

subtract the translation first T·P = (9 − 15 − (−3)) = (88) find T⁻¹: det T = 2(1) − 1(1) = 1 T⁻¹ = (1−1−12) pre-multiply: P = T⁻¹ · (8 ; 8) top: 1(8) + (−1)(8) = 0 bot: −1(8) + 2(8) = 8 P = (0, 8) always subtract translation first, then invert — order matters.

💡 Top tips

Write column vectors first — coordinates as (x, y) cannot be multiplied directly by a 2×2 matrix; they must be stacked vertically as (x ; y).
Use a position matrix P for multi-point shapes — one matrix multiplication T·P gives every image at once.
The GDC is your friend — key in matrices and compute T·P, T⁻¹, Tⁿ directly. Especially helpful when entries are fractions or surds.
Order of operations — multiply by T first, then add the translation. When reversing, subtract translation first, then apply T⁻¹.
Check your answer by transforming forward to make sure you get the stated image (or backward to recover the object).

⚠ Common mistakes

Adding the translation before multiplying — the formula is T·v + t, not T·(v + t). The two give different results.
Multiplying in the wrong order — T·P (matrix times position) is correct; P·T is not even defined for a 2-row P and a 2×2 T in general.
Mixing rows and columns of T — (a b ; c d) means top row is (a, b), bottom row is (c, d). Always row-of-T · column-of-vector.
Forgetting to subtract the translation when going backwards — T⁻¹·(image) gives the wrong answer if a translation was applied.
Reading the GDC display wrong — a value like −(2+√3)/2 is not the same as (−2+√3)/2. Copy fractions and signs carefully.

Next up — Matrices of Geometric Transformations. Same machinery, but now you’ll identify the specific matrices for rotations, reflections, enlargements, and stretches — mostly given in the formula booklet, with a clever way to derive any of them from where (1, 0) and (0, 1) get sent.

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