Maths

Matrices & linear transformations

Edexcel Core Pure CP3

A-level Further Maths (9FM0) · exam-style practice, examiner-report intelligence and the tools that drill it.

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  • Matrices encode linear rules. Add or subtract only matrices of the same order; multiply ABAB only when the number of columns of AA matches the number of rows of BB. In general ABBAAB\ne BA, so order is part of the method (CP-3.1).
  • The zero matrix plays the role of 00 and the identity matrix II plays the role of 11: AI=IA=AAI=IA=A. A transformation followed by its inverse has matrix A1A=IA^{-1}A=I (CP-3.2 and CP-3.6).
  • For a 2-D transformation, the columns of the matrix are the images of (10)\binom10 and (01)\binom01. If transformation AA happens first and BB second, the combined matrix is BABA, because xAxBAx\mathbf{x}\mapsto A\mathbf{x}\mapsto BA\mathbf{x} (CP-3.3).
  • For a 3-D transformation, the three columns are the images of the standard basis vectors (100)\begin{pmatrix}1\\0\\0\end{pmatrix}, (010)\begin{pmatrix}0\\1\\0\end{pmatrix} and (001)\begin{pmatrix}0\\0\\1\end{pmatrix} (CP-3.3).
  • Worked example (CP-3.3): reflection in the plane z=0z=0 maps (x,y,z)(x,y,z) to (x,y,z)(x,y,-z), so its matrix is R=(100010001)R=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}. For example, R(234)=(234)R\begin{pmatrix}2\\-3\\4\end{pmatrix}=\begin{pmatrix}2\\-3\\-4\end{pmatrix}, while R2=IR^2=I confirms that reflecting twice restores every point.
  • Invariant points satisfy Ax=xA\mathbf{x}=\mathbf{x}, so solve (AI)x=0(A-I)\mathbf{x}=\mathbf0. For an invariant line through the origin, transform (x,mx)(x,mx) and require its image to remain on the line. For a general line y=mx+cy=mx+c, transform (x,mx+c)(x,mx+c) and impose Y=mX+cY=mX+c for every xx; if every point is fixed as well as remaining on the line, it is a line of invariant points (CP-3.4).
  • The determinant is a signed scale factor. In 2-D, areas are multiplied by detA|\det A|; in 3-D, volumes are multiplied by detA|\det A|. A negative determinant reverses orientation (CP-3.5).
  • A matrix is singular exactly when its determinant is zero. Then no inverse exists and dimensions have been collapsed. For A=(2111)A=\begin{pmatrix}2&1\\1&1\end{pmatrix}, detA=1\det A=1 and A1=(1112)A^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix} (CP-3.6).
  • To solve Ax=bA\mathbf{x}=\mathbf{b} with a non-singular 3×33\times3 matrix, use x=A1b\mathbf{x}=A^{-1}\mathbf{b}. Keep the exact matrix visible: method marks depend on using the inverse, not merely reporting calculator output (CP-3.7).
  • Three simultaneous linear equations represent three planes. A unique solution is one common point; infinitely many solutions give a common line or coincident plane; no solution means the planes have no common point (CP-3.8).
  • Worked example: the reflection in y=xy=x has matrix (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}. It has determinant 1-1, so it preserves area but reverses orientation; every point on y=xy=x is invariant.
  • Common errors: multiplying in chronological rather than reverse matrix order, using 1/detA1/\det A without the adjugate, ignoring a negative determinant, and writing 'no inverse' without connecting it to a collapsed transformation or dependent equations.
  • Exam technique: calculate the determinant early. It tells you whether inverse methods are legal, predicts the geometry, and often exposes arithmetic errors before they spread.

Where students actually lose marks

State the order of successive transformations in words, then write the product. This prevents the most common matrix-order error.

Original 9FM0-style exam guidance

A determinant question normally needs interpretation as well as calculation: scale factor uses the absolute value, while the sign records orientation.

Original 9FM0-style exam guidance

When an inverse does not exist, translate the row dependence into geometry rather than stopping at 'determinant zero'.

Original 9FM0-style exam guidance

Try it — exam-style

Hard
7 marks
ORIGINAL

Transformation AA has matrix (1201)\begin{pmatrix}1&2\\0&1\end{pmatrix} and is followed by transformation BB with matrix (0110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}. Find the combined matrix and an invariant line through the origin.

Hard
8 marks
ORIGINAL

Use a matrix method to solve x+y+z=6x+y+z=6, 2xy+z=32x-y+z=3, x+2yz=2x+2y-z=2. You must show the inverse matrix used.

Easy
4 marks
ORIGINAL

A linear transformation has matrix M=(2111)M=\begin{pmatrix}-2&1\\1&1\end{pmatrix}. A triangle has area 55. Find the area of its image and state what happens to orientation.

Medium
6 marks
ORIGINAL

Consider x+y+z=1x+y+z=1, 2x+2y+2z=k2x+2y+2z=k and xy=0x-y=0. Describe the geometrical solution set when (a) k=2k=2 and (b) k2k\ne2.

Medium
6 marks
ORIGINAL

The transformation TT has matrix A=(110010001)A=\begin{pmatrix}1&1&0\\0&1&0\\0&0&-1\end{pmatrix}. Find A1A^{-1} and the point whose image under TT is (5,2,3)(5,2,-3).

Questions are written in the style of past Edexcel papers (source shown on each) — never copied from them.

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