Further Mathematics · Pearson Edexcel 9FM0 · Topic 3

Matrices

Grids of numbers that act on vectors. You add and multiply matrices, find determinants and inverses, use them as transformations (rotations, reflections, enlargements), find invariant points and lines, and solve three equations in three unknowns.

  • 8 sub-topics
  • Taught in Year 12
  • 132 practice questions
  • 8 animated lessons
Practice questions (132) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.1Matrix operationsY12
  2. 3.2Zero and identity matricesY12
  3. 3.3Linear transformationsY12
  4. 3.4Invariant points and invariant linesY12
  5. 3.5DeterminantsY12
  6. 3.6Inverse matricesY12
  7. 3.7Solving three linear equationsY12
  8. 3.8Geometry of three planesY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 132 questions in this topic, free, no account needed.

Matrices

Easy

Inverse of a 2 × 2 matrix · Year 12 · 4 marks

A\mathbf{A}=(2153){}= \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}
  1. (a)
    Find det⁡A\det \mathbf{A}.
    (1)
  2. (b)
    Find A−1\mathbf{A}^{-1}.
    (2)
  3. (c)
    A triangle of area 4 square units is transformed by A\mathbf{A}. State the area of its image.
    (1)
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Part (a)1 mark

  1. For (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} the determinant is ad−bcad - bc.det⁡A\det\mathbf{A}=2×3−1×5{}= 2 \times 3 - 1 \times 5=1{}= 1

Part (b)2 marks

  1. Swap the numbers on the leading diagonal, change the signs of the other two, then divide by the determinant.A−1\mathbf{A}^{-1}=11(3−1−52){}= \dfrac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}
  2. Dividing by 1 changes nothing.A−1\mathbf{A}^{-1}=(3−1−52){}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

Part (c)1 mark

  1. The determinant is the area scale factor of the transformation.Area =4×∣det⁡A∣=4= 4 \times |\det\mathbf{A}| = 4 square units

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Matrices

Medium

Solving equations with an inverse · Year 12 · 3 marks

  1. Use (2153)−1\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}^{-1}=(3−1−52){}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} to solve 2x+y=52x + y = 5, 5x+3y=135x + 3y = 13.
    (3)
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  1. Write the equations as Ax=b\mathbf{A}\mathbf{x} = \mathbf{b}, so x\mathbf{x}=A−1b{}= \mathbf{A}^{-1}\mathbf{b}.(xy)\begin{pmatrix} x \\ y \end{pmatrix}=(3−1−52)(513){}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 13 \end{pmatrix}
  2. Multiply out.=(15−13−25+26)= \begin{pmatrix} 15 - 13 \\ -25 + 26 \end{pmatrix}
  3. Check: 4+1=54 + 1 = 5 and 10+3=1310 + 3 = 13.x=2, y=1x = 2,\ y = 1

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Matrices

Hard

Three planes · Year 12 · 5 marks

  1. (a)
    Show that the planes x+y+z=6x + y + z = 6, x+2y+3z=14x + 2y + 3z = 14 and 2x+3y+4z=202x + 3y + 4z = 20 do not meet at a single point.
    (2)
  2. (b)
    Describe how the three planes are arranged.
    (3)
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Part (a)2 marks

  1. If the matrix of coefficients is singular, there is no unique solution.det⁡(111123234)\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}=1(−1)−1(−2)+1(−1){}= 1(-1) - 1(-2) + 1(-1)=0{}= 0
  2. The matrix is singular, so there is no single point of intersection.

Part (b)3 marks

  1. Look for a combination of equations that gives the third.The first equation plus the second gives 2x+3y+4z=202x + 3y + 4z = 20.
  2. The third plane contains every point the first two share.This is the third equation, so the equations are consistent.
  3. Consistent but singular means infinitely many solutions on a line.The planes form a sheaf: they all meet in one common line.

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