Further Mathematics · Pearson Edexcel 9FM0 · Option FP2 · Topic 3

Further matrix algebra

Eigenvalues and eigenvectors, diagonalising matrices, and the Cayley–Hamilton theorem.

  • 3 sub-topics
  • Taught in Year 13
  • 40 practice questions
  • 3 animated lessons

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. 3.1Eigenvalues and eigenvectorsY13
  2. 3.2DiagonalisationY13
  3. 3.3Cayley–Hamilton theoremY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 40 questions in this topic, free, no account needed.

Further matrix algebra

Not needed for Year 12Easy

The characteristic equation · Year 13 · 1 mark

  1. Which equation gives the eigenvalues λ\lambda of a square matrix M\mathbf{M}?
    (1)

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  1. Mv=λv\mathbf{Mv} = \lambda\mathbf{v} has a non-zero solution vv only when M−λI\mathbf{M} - \lambda\mathbf{I} is singular.A: det⁡(M−λI)\det(\mathbf{M} - \lambda\mathbf{I})=0{}= 0

Total1

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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Further matrix algebra

Not needed for Year 12Medium

Eigenvalues and eigenvectors · Year 13 · 4 marks

M\mathbf{M}=(−2503){}= \begin{pmatrix} -2 & 5 \\ 0 & 3 \end{pmatrix}
  1. Find the eigenvalues and corresponding eigenvectors.
    (4)
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  1. det⁡(M−λI)=0\det (M - \lambda I) = 0.(λ+2)(λ−3)=0(\lambda + 2)(\lambda - 3) = 0
  2. λ=−2,3\lambda = -2, 3
  3. λ\lambda=−2:(10){}= -2: \begin{pmatrix} 1 \\ 0 \end{pmatrix}
  4. λ\lambda=3:(11){}= 3: \begin{pmatrix} 1 \\ 1 \end{pmatrix}

Total4

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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Further matrix algebra

Not needed for Year 12Hard

Eigenvalues and eigenvectors · Year 13 · 7 marks

A\mathbf{A}=(−100320−7−24){}= \begin{pmatrix} -1 & 0 & 0 \\ 3 & 2 & 0 \\ -7 & -2 & 4 \end{pmatrix}.
  1. (a)
    Show that −1-1 is an eigenvalue of A\mathbf{A}, and find the other two eigenvalues.
    (5)
  2. (b)
    Find an eigenvector corresponding to the eigenvalue −1-1.
    (2)
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Part (a)5 marks

  1. The characteristic equation.det⁡(A−λI)\det(\mathbf{A} - \lambda\mathbf{I})=0{}= 0
  2. Expand the determinant.−λ3- \lambda^{3}+5λ2{}+ 5 \lambda^{2}−2λ{}- 2 \lambda−8{}- 8=0{}= 0
  3. Substitute to show it is a root.λ=−1: 0=0\lambda = -1: \ 0 = 0
  4. Divide out the factor for λ=−1\lambda = -1 and solve the quadratic.λ=2, 4\lambda = 2, \ 4

Part (b)2 marks

  1. Solve the homogeneous system.(A+I)x(\mathbf{A} + \mathbf{I})\mathbf{x}=0{}= \mathbf{0}
  2. Any non-zero multiple.(1−11)\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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