Further Maths Linear transformations Practice Questions

Free Further Maths Linear transformations practice questions with full step-by-step worked solutions. Covers linear-transformations, rotation, matrix, reflection. Practise exam-style problems and check your method.

linear-transformationsrotationmatrixreflectionenlargementstretch
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Write down the matrix that represents a rotation of 90∘90^\circ anticlockwise about the origin.
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Worked solution

  1. Write down the general rotation matrix

    (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

    An anticlockwise rotation through θ\theta has this matrix.

  2. Substitute θ=90∘\theta=90^\circ

    cos⁡90∘=0,sin⁡90∘=1\cos 90^\circ=0,\quad \sin 90^\circ=1

    Evaluate the trigonometric functions exactly.

  3. State the rotation matrix

    (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

    This matrix rotates every point about the origin.

Answer
(0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
Question 2
2 markseasy
Describe fully the single transformation represented by the matrix M=(4004)\mathbf{M}=\begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}.
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Worked solution

  1. Find the images of the base vectors

    (4004)(10)=(40),(4004)(01)=(04)\begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 4 \\ 0 \end{pmatrix},\quad \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 4 \end{pmatrix}

    The columns of the matrix tell you where (1,0)(1,0) and (0,1)(0,1) go.

  2. Find the determinant

    det⁡M=16\det \mathbf{M}=16

    The sign and size of the determinant narrow down the type of transformation.

  3. Select the correct description

    Enlargement, centre the origin, scale factor 44

    This description matches the matrix exactly.

Answer
Enlargement, centre the origin, scale factor 44
Question 3
3 marksintermediate
Describe fully the single transformation represented by the matrix M=(2001)\mathbf{M}=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}.
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Worked solution

  1. Find the images of the base vectors

    (2001)(10)=(20),(2001)(01)=(01)\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 2 \\ 0 \end{pmatrix},\quad \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}

    The columns of the matrix tell you where (1,0)(1,0) and (0,1)(0,1) go.

  2. Find the determinant

    det⁡M=2\det \mathbf{M}=2

    The sign and size of the determinant narrow down the type of transformation.

  3. Compare with the standard forms

    (2001)\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}

    Match the matrix against the rotation, reflection, enlargement and stretch templates.

  4. Check whether lengths are preserved

    ∣M(10)∣=2\left|\mathbf{M}\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right|=2

    Rotations and reflections preserve length; enlargements and stretches do not.

  5. Test a second point

    (2001)(11)=(21)\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 2 \\ 1 \end{pmatrix}

    A second image confirms the transformation you have identified.

  6. Select the correct description

    Stretch parallel to the xx-axis, scale factor 22

    This description matches the matrix exactly.

Answer
Stretch parallel to the xx-axis, scale factor 22
Question 4
5 markshard
Which of the following matrices represents a reflection in the line y=3xy=\sqrt{3}x?
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Worked solution

  1. Recall the standard matrix for this transformation

    (−12323212)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}

    Build the matrix from the images of (1,0)(1,0) and (0,1)(0,1).

  2. Find the images of the base vectors

    (−12323212)(10)=(−1232),(−12323212)(01)=(3212)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix},\quad \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}

    The columns of the matrix tell you where (1,0)(1,0) and (0,1)(0,1) go.

  3. Find the determinant

    det⁡M=−1\det \mathbf{M}=-1

    The sign and size of the determinant narrow down the type of transformation.

  4. Check whether lengths are preserved

    ∣M(10)∣=1\left|\mathbf{M}\begin{pmatrix} 1 \\ 0 \end{pmatrix}\right|=1

    Rotations and reflections preserve length; enlargements and stretches do not.

  5. Test a second point

    (−12323212)(11)=(−12+3212+32)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -\frac{1}{2} + \frac{\sqrt{3}}{2} \\ \frac{1}{2} + \frac{\sqrt{3}}{2} \end{pmatrix}

    A second image confirms the transformation you have identified.

  6. Check the image of the point (2,0)(2,0)

    (−12323212)(20)=(−13)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} 2 \\ 0 \end{pmatrix}=\begin{pmatrix} -1 \\ \sqrt{3} \end{pmatrix}

    A further image confirms the description, including any scale factor.

  7. Recall the general rotation matrix

    (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

    An anticlockwise rotation through θ\theta about the origin has this matrix.

  8. Recall the general reflection matrix

    (cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}

    A reflection in the line through the origin at angle θ\theta has this matrix.

  9. Recall the effect on the base vectors

    M(10)=column 1,M(01)=column 2\mathbf{M}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\text{column }1,\quad \mathbf{M}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\text{column }2

    The columns of the matrix are the images of (1,0)(1,0) and (0,1)(0,1).

  10. Select the correct matrix

    (−12323212)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}

    This matrix carries out the stated transformation.

Answer
(−12323212)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}
Question 5
8 markschallenging
Which of the following lines is an invariant line of the transformation represented by M=(3143)\mathbf{M}=\begin{pmatrix} 3 & 1 \\ 4 & 3 \end{pmatrix}?
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Worked solution

  1. Substitute a general point of y=mxy=mx

    (3143)(xmx)=((3+1m)x(4+3m)x)\begin{pmatrix} 3 & 1 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix}=\begin{pmatrix} (3+1m)x \\ (4+3m)x \end{pmatrix}

    The image of a point on the line must lie on the same line.

  2. Form the quadratic in mm

    m2−4=0m^{2} -4=0

    This is bm2+(a−d)m−c=0bm^2+(a-d)m-c=0.

  3. Solve for the gradients

    m=−2, 2m=-2,\ 2

    Each root gives an invariant line through the origin.

  4. Check the listed option

    (3143)(1−2)=(1−2)\begin{pmatrix} 3 & 1 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix}=\begin{pmatrix} 1 \\ -2 \end{pmatrix}

    The sample point maps to another point of the same line.

  5. Recall the general rotation matrix

    (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

    An anticlockwise rotation through θ\theta about the origin has this matrix.

  6. Recall the general reflection matrix

    (cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}

    A reflection in the line through the origin at angle θ\theta has this matrix.

  7. Recall the effect on the base vectors

    M(10)=column 1,M(01)=column 2\mathbf{M}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\text{column }1,\quad \mathbf{M}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\text{column }2

    The columns of the matrix are the images of (1,0)(1,0) and (0,1)(0,1).

  8. Recall the area scale factor

    area factor=∣det⁡M∣\text{area factor}=\left|\det \mathbf{M}\right|

    The modulus of the determinant scales areas under the transformation.

  9. Recall the enlargement matrix

    (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}

    An enlargement centre the origin with scale factor kk is a scalar matrix.

  10. Recall the condition for an invariant point

    Mv=v\mathbf{M}\mathbf{v}=\mathbf{v}

    An invariant point is mapped to itself by the transformation.

  11. Recall the condition for an invariant line through the origin

    bm2+(a−d)m−c=0bm^2+(a-d)m-c=0

    Substituting (x,mx)(x,mx) into M\mathbf{M} and demanding the image satisfies y=mxy=mx gives this quadratic in mm.

  12. Recall that the origin is always invariant

    M(00)=(00)\mathbf{M}\begin{pmatrix} 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}

    Every linear transformation fixes the origin.

  13. Recall the order of composition

    first A, then B ⇒ BA\text{first }\mathbf{A}\text{, then }\mathbf{B}\ \Rightarrow\ \mathbf{B}\mathbf{A}

    The matrix of the second transformation is written on the left.

  14. Recall the inverse of a 2×22\times 2 matrix

    M−1=1ad−bc(d−b−ca)\mathbf{M}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

    Swap the leading diagonal, negate the other diagonal, divide by the determinant.

  15. Select the invariant line

    y=−2xy=-2x

    Points of this line are mapped onto the same line.

Answer
y=−2xy=-2x

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