Further matrix algebra Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Further matrix algebra questions. See exactly how to solve problems on characteristic-equation, determinant, eigenvalues, eigenvectors.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Find the characteristic equation of A=(4123)\mathbf{A} = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}, giving your answer in the form λ2+aλ+b=0\lambda^{2}+a\lambda+b = 0.

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(4123)−λ(1001)=(4−λ123−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}4 - \lambda & 1 \\ 2 & 3 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣4−λ123−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}4 - \lambda & 1 \\ 2 & 3 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Expand the 2×22\times2 determinant

    (4−λ)(3−λ)−(1)(2)=0\left(4-\lambda\right)\left(3-\lambda\right) - \left(1\right)\left(2\right) = 0

    Use ad−bcad-bc on the matrix A−λI\mathbf{A}-\lambda\mathbf{I}.

  4. State the characteristic equation

    λ2−7λ+10=0\lambda^{2} - 7 \lambda + 10 = 0

    This is the required equation, written with leading coefficient 11.

Answer
λ2−7λ+10=0\lambda^{2} - 7 \lambda + 10 = 0
Question 2
2 markseasy
Find the characteristic equation of A=(1221)\mathbf{A} = \begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix}, giving your answer in the form λ2+aλ+b=0\lambda^{2}+a\lambda+b = 0.

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(1221)−λ(1001)=(1−λ221−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 - \lambda & 2 \\ 2 & 1 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣1−λ221−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}1 - \lambda & 2 \\ 2 & 1 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Expand the 2×22\times2 determinant

    (1−λ)(1−λ)−(2)(2)=0\left(1-\lambda\right)\left(1-\lambda\right) - \left(2\right)\left(2\right) = 0

    Use ad−bcad-bc on the matrix A−λI\mathbf{A}-\lambda\mathbf{I}.

  4. State the characteristic equation

    λ2−2λ−3=0\lambda^{2} - 2 \lambda - 3 = 0

    This is the required equation, written with leading coefficient 11.

Answer
λ2−2λ−3=0\lambda^{2} - 2 \lambda - 3 = 0
Question 3
2 markseasy
Find the characteristic equation of A=(5412)\mathbf{A} = \begin{pmatrix}5 & 4 \\ 1 & 2\end{pmatrix}, giving your answer in the form λ2+aλ+b=0\lambda^{2}+a\lambda+b = 0.

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(5412)−λ(1001)=(5−λ412−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}5 & 4 \\ 1 & 2\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}5 - \lambda & 4 \\ 1 & 2 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣5−λ412−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}5 - \lambda & 4 \\ 1 & 2 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Expand the 2×22\times2 determinant

    (5−λ)(2−λ)−(4)(1)=0\left(5-\lambda\right)\left(2-\lambda\right) - \left(4\right)\left(1\right) = 0

    Use ad−bcad-bc on the matrix A−λI\mathbf{A}-\lambda\mathbf{I}.

  4. State the characteristic equation

    λ2−7λ+6=0\lambda^{2} - 7 \lambda + 6 = 0

    This is the required equation, written with leading coefficient 11.

Answer
λ2−7λ+6=0\lambda^{2} - 7 \lambda + 6 = 0
Question 4
2 markseasy
Find the eigenvalues of A=(3102)\mathbf{A} = \begin{pmatrix}3 & 1 \\ 0 & 2\end{pmatrix}, giving them in increasing order.

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(3102)−λ(1001)=(3−λ102−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}3 & 1 \\ 0 & 2\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}3 - \lambda & 1 \\ 0 & 2 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣3−λ102−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}3 - \lambda & 1 \\ 0 & 2 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Factorise the characteristic polynomial

    λ2−5λ+6=(λ−3)(λ−2)=0\lambda^{2} - 5 \lambda + 6 = \left(\lambda - 3\right) \left(\lambda - 2\right) = 0

    Look for integer roots among the factors of the constant term.

  4. State the eigenvalues in increasing order

    λ=2, 3\lambda = 2,\ 3

    These are the 22 eigenvalues of A\mathbf{A}.

Answer
λ=2, 3\lambda = 2,\ 3
Question 5
2 markseasy
Find the eigenvalues of A=(4123)\mathbf{A} = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}, giving them in increasing order.

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(4123)−λ(1001)=(4−λ123−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}4 - \lambda & 1 \\ 2 & 3 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣4−λ123−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}4 - \lambda & 1 \\ 2 & 3 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Factorise the characteristic polynomial

    λ2−7λ+10=(λ−5)(λ−2)=0\lambda^{2} - 7 \lambda + 10 = \left(\lambda - 5\right) \left(\lambda - 2\right) = 0

    Look for integer roots among the factors of the constant term.

  4. State the eigenvalues in increasing order

    λ=2, 5\lambda = 2,\ 5

    These are the 22 eigenvalues of A\mathbf{A}.

Answer
λ=2, 5\lambda = 2,\ 5

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