Diagonalisation and the Cayley-Hamilton theorem

Diagonalisation and the Cayley-Hamilton theorem Quiz

Spec 3.2–3 · Edexcel A-Level Further Maths: Quiz

Studying this course? Keep your progress in one place.

Pick an answer for each. Quizzes never use energy.

5 questions, one answer each

Each answer is checked as you go, with why.

See all 5 questions

  1. 1. The columns of P\mathbf{P} are eigenvectors of A\mathbf{A} for λ=3\lambda=3 then λ=7\lambda=7. Then P−1AP\mathbf{P}^{-1}\mathbf{A}\mathbf{P} is

    • A. diag(7,3)\mathrm{diag}(7,3)
    • B. (3700)\begin{pmatrix} 3 & 7 \\ 0 & 0 \end{pmatrix}
    • C. diag(3,7)\mathrm{diag}(3,7)
    • D. diag(10,21)\mathrm{diag}(10,21)

  2. 2. Which matrix can be orthogonally diagonalised?

    • A. (1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}
    • B. (1223)\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}
    • C. (01−10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
    • D. (1101)\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}

  3. 3. A\mathbf{A} has characteristic equation λ2−3λ+2=0\lambda^2-3\lambda+2=0. By Cayley-Hamilton, A2=\mathbf{A}^2=

    • A. 3A−2I3\mathbf{A}-2\mathbf{I}
    • B. 3A+2I3\mathbf{A}+2\mathbf{I}
    • C. 2I−3A2\mathbf{I}-3\mathbf{A}
    • D. 3A−23\mathbf{A}-2

  4. 4. For an orthogonal matrix Q\mathbf{Q}, Q−1\mathbf{Q}^{-1} equals

    • A. Q2\mathbf{Q}^2
    • B. −Q-\mathbf{Q}
    • C. det⁡Q\det\mathbf{Q}
    • D. QT\mathbf{Q}^{\mathrm{T}}

  5. 5. A\mathbf{A} has eigenvalues 22 and 33. The eigenvalues of A3\mathbf{A}^3 are

    • A. 66 and 99
    • B. 88 and 2727
    • C. 55 and 66
    • D. 22 and 33

Quiz