Further Maths Further matrix algebra Practice Questions

Free Further Maths Further matrix algebra practice questions with full step-by-step worked solutions. Covers characteristic-equation, determinant, eigenvalues, eigenvectors. Practise exam-style problems and check your method.

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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.
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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
Given A=(2−1−12)\mathbf{A} = \begin{pmatrix}2 & -1 \\ -1 & 2\end{pmatrix}, which of the following gives the sum and the product of the eigenvalues of A\mathbf{A}, in that order?
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Worked solution

  1. Recall the two relations

    ∑λi=tr⁡A,∏λi=det⁡A\sum\lambda_{i} = \operatorname{tr}\mathbf{A},\qquad \prod\lambda_{i} = \det\mathbf{A}

    They come from comparing coefficients in det⁡(λI−A)=∏(λ−λi)\det\left(\lambda\mathbf{I}-\mathbf{A}\right) = \prod\left(\lambda-\lambda_{i}\right).

  2. Read off the trace

    tr⁡A=2+2=4\operatorname{tr}\mathbf{A} = 2 + 2 = 4

    The trace is the sum of the leading diagonal.

  3. Evaluate the determinant

    det⁡A=∣2−1−12∣=3\det\mathbf{A} = \begin{vmatrix}2 & -1 \\ -1 & 2\end{vmatrix} = 3

    The determinant is computed in the usual way.

  4. Select the correct pair

    sum=4,product=3\text{sum} = 4,\qquad \text{product} = 3

    The sum of the eigenvalues is the trace and their product is the determinant.

Answer
44 and 33
Question 3
4 marksintermediate
Which of the following is the normalised eigenvector of A=(6−2−29)\mathbf{A} = \begin{pmatrix}6 & -2 \\ -2 & 9\end{pmatrix} corresponding to the eigenvalue λ=5\lambda = 5, with a positive first component?
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Worked solution

  1. Substitute λ=5\lambda = 5 into A−λI\mathbf{A}-\lambda\mathbf{I}

    A−(5)I=(1−2−24)\mathbf{A}-\left(5\right)\mathbf{I} = \begin{pmatrix}1 & -2 \\ -2 & 4\end{pmatrix}

    The eigenvector for this eigenvalue is a non-zero solution of (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0}.

  2. Write out the simultaneous equations

    (1)x+(−2)y=0,(−2)x+(4)y=0\left(1\right)x + \left(-2\right)y = 0,\qquad \left(-2\right)x + \left(4\right)y = 0

    This system is singular by construction, so it has infinitely many solutions.

  3. Solve for the ratio of the components

    y=12xy = \frac{1}{2}x

    Only the direction matters, so one component may be chosen freely.

  4. Normalise the eigenvector

    ∣v∣=5⇒v^=(25555)\left|\mathbf{v}\right| = \sqrt{5} \Rightarrow \hat{\mathbf{v}} = \begin{pmatrix}\frac{2 \sqrt{5}}{5} \\ \frac{\sqrt{5}}{5}\end{pmatrix}

    Divide each component by the magnitude of the eigenvector.

  5. Verify the eigenvector directly

    (6−2−29)(21)=(105)=(5)(21)\begin{pmatrix}6 & -2 \\ -2 & 9\end{pmatrix}\begin{pmatrix}2 \\ 1\end{pmatrix} = \begin{pmatrix}10 \\ 5\end{pmatrix} = \left(5\right)\begin{pmatrix}2 \\ 1\end{pmatrix}

    Multiplying by A\mathbf{A} scales the vector by 55, as required.

  6. Note that the rows of A−λI\mathbf{A}-\lambda\mathbf{I} are dependent

    det⁡(A−(5)I)=0\det\left(\mathbf{A}-\left(5\right)\mathbf{I}\right) = 0

    That is exactly why a non-zero solution exists.

  7. Select the normalised eigenvector

    (25555)\begin{pmatrix}\frac{2 \sqrt{5}}{5} \\ \frac{\sqrt{5}}{5}\end{pmatrix}

    This is the unit vector in the direction of the eigenvector for λ=5\lambda = 5, with a positive first component.

Answer
(25555)\begin{pmatrix}\frac{2 \sqrt{5}}{5} \\ \frac{\sqrt{5}}{5}\end{pmatrix}
Question 4
6 markshard
Which of the following matrices can be diagonalised?
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Worked solution

  1. State the test for diagonalisability

    n linearly independent eigenvectors are neededn \text{ linearly independent eigenvectors are needed}

    A 2×22\times2 matrix must supply two independent eigenvectors.

  2. Note the shortcut for distinct eigenvalues

    λ1≠λ2⇒diagonalisable\lambda_{1} \neq \lambda_{2} \Rightarrow \text{diagonalisable}

    Eigenvectors for distinct eigenvalues are always independent.

  3. Find the eigenvalues of the correct option

    (λ−20)(λ−5)=0⇒λ=5, 20\left(\lambda - 20\right) \left(\lambda - 5\right) = 0 \Rightarrow \lambda = 5,\ 20

    Two distinct eigenvalues, so this matrix is diagonalisable.

  4. Examine the other options

    (λ−3)2⇒λ=3, 3,(λ−2)2⇒λ=2, 2\left(\lambda - 3\right)^{2} \Rightarrow \lambda = 3,\ 3,\qquad \left(\lambda - 2\right)^{2} \Rightarrow \lambda = 2,\ 2

    Each of these has a repeated eigenvalue with only one independent eigenvector.

  5. Count the eigenvectors of a rejected option

    rank⁡(B−λI)=1⇒one eigenvector\operatorname{rank}\left(\mathbf{B}-\lambda\mathbf{I}\right) = 1 \Rightarrow \text{one eigenvector}

    A repeated eigenvalue with a rank-11 matrix leaves a one-dimensional eigenspace.

  6. Note that a repeated eigenvalue is not automatically fatal

    B=c I⇒already diagonal\mathbf{B} = c\,\mathbf{I} \Rightarrow \text{already diagonal}

    A repeated eigenvalue only blocks diagonalisation when it is short of eigenvectors.

  7. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  8. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  9. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  10. Recall the shortcut for a 2×22\times2 characteristic equation

    λ2−(tr⁡A)λ+det⁡A=0\lambda^{2} - \left(\operatorname{tr}\mathbf{A}\right)\lambda + \det\mathbf{A} = 0

    The coefficients are the trace and the determinant, which is a quick check on the algebra.

  11. Select the diagonalisable matrix

    (8−6−617)\begin{pmatrix}8 & -6 \\ -6 & 17\end{pmatrix}

    It has two distinct eigenvalues, so it has two independent eigenvectors.

Answer
(8−6−617)\begin{pmatrix}8 & -6 \\ -6 & 17\end{pmatrix}
Question 5
9 markschallenging
The matrix M=(5222)\mathbf{M} = \begin{pmatrix}5 & 2 \\ 2 & 2\end{pmatrix} has λ=1\lambda = 1 as an eigenvalue. Which of the following describes the set of points r\mathbf{r} for which Mr=r\mathbf{M}\mathbf{r} = \mathbf{r}?
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Worked solution

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

    A−λI=(5222)−λ(1001)=(5−λ222−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}5 & 2 \\ 2 & 2\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}5 - \lambda & 2 \\ 2 & 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−λ222−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}5 - \lambda & 2 \\ 2 & 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)−(2)(2)=0\left(5-\lambda\right)\left(2-\lambda\right) - \left(2\right)\left(2\right) = 0

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

  4. Multiply out and collect the terms

    det⁡(A−λI)=λ2−7λ+6\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \lambda^{2} - 7 \lambda + 6

    This is the characteristic polynomial in its expanded form.

  5. Interpret Mr=r\mathbf{M}\mathbf{r} = \mathbf{r}

    Mr=r=1⋅r\mathbf{M}\mathbf{r} = \mathbf{r} = 1\cdot\mathbf{r}

    A fixed point is an eigenvector with eigenvalue 11 (or the origin).

  6. Find the eigenvector for λ=1\lambda = 1

    (M−I)v=0⇒v=(1−2)\left(\mathbf{M}-\mathbf{I}\right)\mathbf{v} = \mathbf{0} \Rightarrow \mathbf{v} = \begin{pmatrix}1 \\ -2\end{pmatrix}

    Every multiple of this vector is fixed by M\mathbf{M}.

  7. Convert the direction into a line

    y=−2xy = -2x

    The whole line through the origin in this direction consists of invariant points.

  8. Test a point of the line

    M(1−2)=(1−2)\mathbf{M}\begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}1 \\ -2\end{pmatrix}

    The point is mapped to itself, confirming that it is invariant.

  9. Contrast with the other eigenvalue

    λ=6⇒the line y=12x is invariant but its points are not fixed\lambda = 6 \Rightarrow \text{the line } y = \frac{1}{2}x \text{ is invariant but its points are not fixed}

    Points on that line are scaled by 66, so only the origin stays put.

  10. Note that the origin alone is not the answer

    λ=1 is an eigenvalue⇒a whole line is fixed\lambda = 1 \text{ is an eigenvalue} \Rightarrow \text{a whole line is fixed}

    A non-trivial eigenvector for λ=1\lambda = 1 exists, so more than the origin is fixed.

  11. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  12. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  13. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  14. Recall the shortcut for a 2×22\times2 characteristic equation

    λ2−(tr⁡A)λ+det⁡A=0\lambda^{2} - \left(\operatorname{tr}\mathbf{A}\right)\lambda + \det\mathbf{A} = 0

    The coefficients are the trace and the determinant, which is a quick check on the algebra.

  15. Select the correct description

    y=−2xy = -2x

    This is the line of invariant points of the transformation.

Answer
The line of invariant points y=−2xy = -2x

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