Further Mathematics · AQA 7367 · Topic C

Matrices

Matrices as transformations, determinants, inverses, solving equations in three unknowns, and eigenvalues, eigenvectors and diagonalisation.

  • 11 sub-topics
  • 4 in Year 12, 5 in Year 13
  • 169 practice questions
  • 11 animated lessons
Practice questions (169) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA 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. C1Adding, subtracting and multiplying matricesY12
  2. C2Zero and identity matricesY12
  3. C3Matrices as linear transformations in 2D and 3DY12
  4. C4Invariant points and invariant linesY12
  5. C5Determinants as area and volume scale factorsY13
  6. C6Singular matrices and inverse matricesY13
  7. C7Solve three linear simultaneous equations using the inverse matrixY13
  8. C8The geometry of three simultaneous equationsY13
  9. C9Factorise determinants using row and column operationsY13
  10. C10Eigenvalues, eigenvectors and the characteristic equationY13
  11. C11Diagonalising a matrix and finding its powersY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 169 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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