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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn7367Paper 1 · Compulsory contentMatricesTransformations, invariant points and lines
Matrices · Subtopic 2 of 3

Matrix Transformations and Invariance

The columns of a transformation matrix are the images of the unit vectors i\mathbf i and j\mathbf j. Rotations, reflections, enlargements and stretches all have standard matrices. BA\mathbf{BA} means "do A\mathbf A, then B\mathbf B". Invariant points satisfy Mx=x\mathbf M\mathbf x=\mathbf x; invariant lines map onto themselves.

Cambridge 9231 Paper 1Edexcel YFM01 WFM01Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Find the matrix of a transformation from the images of i\mathbf i and j\mathbf j.
  • Use standard matrices for rotations, reflections, enlargements and stretches in 2D, and rotations about the axes in 3D.
  • Combine transformations, remembering BA\mathbf{BA} means A\mathbf A first.
  • Find invariant points and invariant lines.
Pause and think

A reflection in the line y=xy=x swaps the coordinates of every point. Which points don't move?

Section 2 of 14

Core idea

Start hereWhere do the corners go?

Every rotation, reflection or stretch of the plane that keeps the origin fixed is completely decided by just two arrows: where (1,0)(1,0) goes and where (0,1)(0,1) goes. Write those two images side by side and you have the transformation's matrix.

The big idea

Columns of M\mathbf M = images of i\mathbf i, j\mathbf j (k\mathbf k). "A\mathbf A then B\mathbf B" is BA\mathbf{BA}. Invariant points: Mx=x\mathbf M\mathbf x=\mathbf x. Invariant lines map onto themselves.

1. Building matrices from unit vectors

Why the columns are the images
  1. (abcd)(10)=(ac)\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a\\c\end{pmatrix}
    Picks out column 1.
  2. (abcd)(01)=(bd)\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}b\\d\end{pmatrix}
    Picks out column 2.
  3. M(xy)=x(ac)+y(bd)\mathbf M\begin{pmatrix}x\\y\end{pmatrix}=x\begin{pmatrix}a\\c\end{pmatrix}+y\begin{pmatrix}b\\d\end{pmatrix}
    Every vector is xi+yjx\mathbf i+y\mathbf j, so its image is fixed by these two.
The rotation matrix
  1. i↦(cos⁡θ,sin⁡θ)\mathbf i\mapsto(\cos\theta,\sin\theta)
    Rotate (1,0)(1,0) anticlockwise by θ\theta round the unit circle.
  2. j↦(−sin⁡θ,cos⁡θ)\mathbf j\mapsto(-\sin\theta,\cos\theta)
    (0,1)(0,1) starts 90∘90^\circ further round.
  3. Rθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\mathbf R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}
    Write them as columns.
Reflection in y=xtan⁡θ: (cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\text{Reflection in }y=x\tan\theta:\ \begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}
where
θ\theta
angle of the mirror line with the xx-axis.
det⁡=−1\det=-1
reflections reverse orientation.
Key takeaway3D rotations Rotation about the zz-axis: (cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001)\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}; the axis's own unit vector stays fixed. Similarly for the xx- and yy-axes.

2. Combining transformations

Why "A\mathbf A then B\mathbf B" is BA\mathbf{BA}
  1. x → A  Ax\mathbf x\ \xrightarrow{\ \mathbf A\ }\ \mathbf{Ax}
    First transformation.
  2. Ax → B  B(Ax)=(BA)x\mathbf{Ax}\ \xrightarrow{\ \mathbf B\ }\ \mathbf B(\mathbf{Ax})=(\mathbf{BA})\mathbf x
    Second acts on the result; associativity.
Worked exampleReflect then rotate

Reflection in the xx-axis, then rotation 90∘90^\circ anticlockwise about OO. Single transformation?

Goal
Matrix and description.
Strategy
BA\mathbf{BA}.
  1. A=(100−1), B=(0−110)\mathbf A=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\ \mathbf B=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
    Standard matrices.
  2. BA=(0110)\mathbf{BA}=\begin{pmatrix}0&1\\1&0\end{pmatrix}
    Multiply in the right order.
AnswerReflection in y=xy=x.
✓ Check(1,0)→(1,0)→(0,1)(1,0)\to(1,0)\to(0,1); reflection in y=xy=x also sends (1,0)(1,0) to (0,1)(0,1) ✓.

3. Invariant points and lines

Invariant lines through the origin
  1. (abcd)(xmx)=((a+bm)x(c+dm)x)\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\mx\end{pmatrix}=\begin{pmatrix}(a+bm)x\\(c+dm)x\end{pmatrix}
    Image of a general point on y=mxy=mx.
  2. c+dm=m(a+bm)c+dm=m(a+bm)
    The image must also lie on y=mxy=mx.
  3. bm2+(a−d)m−c=0bm^2+(a-d)m-c=0
    A quadratic in mm: up to two invariant lines through OO.
For lines not through OO, use y=mx+cy=mx+c and compare both gradient and intercept.
Worked exampleM=(3122)\mathbf M=\begin{pmatrix}3&1\\2&2\end{pmatrix}

Find its invariant lines through OO.

Goal
Gradients.
Strategy
Quadratic above.
  1. m2+(3−2)m−2=0m^2+(3-2)m-2=0
    b=1b=1, a−d=1a-d=1, c=2c=2.
  2. (m−1)(m+2)=0(m-1)(m+2)=0
    Factorise.
Answery=xy=x and y=−2xy=-2x.
✓ Check(1,1)→(4,4)(1,1)\to(4,4) on y=xy=x ✓; (1,−2)→(1,−2)(1,-2)\to(1,-2) on y=−2xy=-2x ✓ (in fact a line of invariant points).
Common misconceptionInvariant line vs line of invariant points On an invariant line, points may slide along it; on a line of invariant points, every point stays put.
Pause and think

Which transformation has every line through OO invariant?

Summary

  • Columns = images of unit vectors.
  • BA\mathbf{BA} for "A\mathbf A then B\mathbf B".
  • det⁡\det: area factor; negative = reflection-type.
  • Invariant lines from a quadratic in mm.
Key words
Linear transformation
Map preserving addition and scaling; fixes the origin.
Invariant line
Line mapped onto itself.
HistoryLinear algebra. Hermann Grassmann's 1844 work on "extension" treated geometry with linear transformations. It was so ahead of its time that it was largely ignored for decades.
Fun factPhone screens. Rotating your phone triggers a 90∘90^\circ rotation matrix applied to every element on screen.
Section 3 of 14

Key formulae

Rotation by θ anticlockwise(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}
Reflection in y = (tan θ)x(cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}
Enlargement / stretch(k00k),(k001) (x-direction)\begin{pmatrix}k&0\\0&k\end{pmatrix},\qquad\begin{pmatrix}k&0\\0&1\end{pmatrix}\ (x\text{-direction})
InvarianceMx=x (points);y=mx+c↦y=mx+c (lines)\mathbf M\mathbf x=\mathbf x\ (\text{points});\qquad y=mx+c\mapsto y=mx+c\ (\text{lines})
TrickWhere do i and j go? Sketch the unit square, draw where (1,0)(1,0) and (0,1)(0,1) end up, and write those two vectors as columns. That's the matrix, every time.
Section 4 of 14

Worked examples

Example 1 · Combined transformation0 / 3 steps
A\mathbf A: reflection in the xx-axis. B\mathbf B: rotation 90∘90^\circ anticlockwise about OO. Find the single matrix for "A\mathbf A then B\mathbf B".
    Example 2 · Invariant lines0 / 3 steps
    Find the invariant lines through the origin of M=(3122)\mathbf M=\begin{pmatrix}3&1\\2&2\end{pmatrix}.
      Section 5 of 14

      Explore the graph

      Rotation matrix
      Loading the interactive graph…
      Try this
      1. The two points are the images of i\mathbf i and j\mathbf j: the columns of the rotation matrix.
      2. Set t=π2t=\frac\pi2: i→(0,1)\mathbf i\to(0,1), j→(−1,0)\mathbf j\to(-1,0).

      Columns (cos⁡t,sin⁡t)(\cos t,\sin t) and (−sin⁡t,cos⁡t)(-\sin t,\cos t).

      In the real worldComputer graphics. Graphics cards chain transformations by multiplying matrices once, then apply the single product to every vertex, which is far faster than applying each transformation separately.
      Section 6 of 14

      Common mistakes

      Incorrect"A\mathbf A then B\mathbf B" is AB\mathbf{AB}
      CorrectBA\mathbf{BA}
      The matrix nearest the vector acts first.
      IncorrectInvariant line = line of invariant points
      CorrectPoints may move along an invariant line
      Only the line as a whole must map to itself.
      IncorrectRotation clockwise with the anticlockwise matrix
      CorrectUse −θ-\theta for clockwise
      The standard matrix is anticlockwise.

      Exam: "A followed by B"

      Me, wisely: so that's BA.

      Section 7 of 14

      Exam tip

      Exam tipDescribe fully. "Rotation" alone scores nothing. Give the angle, the direction and the centre (or axis in 3D).
      Section 8 of 14

      Guided practice

      SA-FP-MTR-G011 markL1 FoundationNon-calculatorStandard matrices
      The matrix (3003)\begin{pmatrix}3&0\\0&3\end{pmatrix} is an enlargement about OO. Find its area scale factor.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-MTR-0012 marksL2 StandardNon-calculatorStandard matrices
      The matrix (12−323212)\begin{pmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{pmatrix} is a rotation about OO. Find the anticlockwise angle in degrees (0≤θ<3600\le\theta<360).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-MTR-0022 marksL2 StandardNon-calculatorCombining
      P=(2001)\mathbf P=\begin{pmatrix}2&0\\0&1\end{pmatrix} (stretch) and Q=(0−110)\mathbf Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix} (rotation). The point (1,3)(1,3) undergoes P\mathbf P then Q\mathbf Q. Find the xx-coordinate of its image.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-MTR-0033 marksL3 ExamNon-calculatorInvariance
      Find the gradients of the invariant lines through OO of (3122)\begin{pmatrix}3&1\\2&2\end{pmatrix}.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-MTR-0043 marksL3 ExamNon-calculatorInvariance
      M=(2112)\mathbf M=\begin{pmatrix}2&1\\1&2\end{pmatrix} has a line of invariant points y=kxy=kx. Find kk.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-MTR-0053 marksL4 ChallengingNon-calculatorStandard matrices
      The matrix (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix} is a reflection in y=(tan⁡θ)xy=(\tan\theta)x. Find θ\theta in degrees (0≤θ<1800\le\theta<180).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Columns = images of i\mathbf i, j\mathbf j (and k\mathbf k).
      • BA\mathbf{BA}: A\mathbf A first.
      • Invariant points: Mx=x\mathbf M\mathbf x=\mathbf x.
      • Invariant lines: substitute y=mx(+c)y=mx(+c).
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-MTR-Q011 markL1 FoundationNon-calculatorCombining
      Transformation A\mathbf A is followed by B\mathbf B. The single matrix is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      7367 A-level Further Mathematics
      Where
      Paper 1 and 2 · Matrices
      Expected
      Transformations in 2D and 3D, combinations, invariant points and lines (including lines of invariant points).
      Calculator
      Permitted on all papers.
      Typical style
      AQA mixes short multiple-choice questions with longer structured ones.
      Open in
      7367 · Paper 17367 · Paper 2
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).