Hard Further Maths Linear transformations Questions

Challenging, exam-style Further Maths Linear transformations questions with worked solutions. Stretch yourself on the hardest linear-transformations, rotation, matrix, reflection problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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
Question 2
8 markschallenging
The transformation PP is represented by A=(0−110)\mathbf{A}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} and the transformation QQ is represented by B=(−100−1)\mathbf{B}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. Describe fully the single transformation equivalent to PP followed by QQ.
Show worked solution

Worked solution

  1. Identify the order of the product

    C=BA\mathbf{C}=\mathbf{B}\mathbf{A}

    The matrix of the first transformation goes on the right.

  2. Write down the product to be evaluated

    (−100−1)(0−110)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

    Multiply B\mathbf{B} on the left of A\mathbf{A}.

  3. Multiply the matrices

    (−100−1)(0−110)=(01−10)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

    Use the row-by-column rule for each of the four entries.

  4. Compute the top-left entry

    −1×0+0×1=0-1\times 0+0\times 1=0

    Row one of B\mathbf{B} times column one of A\mathbf{A}.

  5. Compute the top-right entry

    −1×−1+0×0=1-1\times -1+0\times 0=1

    Row one of B\mathbf{B} times column two of A\mathbf{A}.

  6. Compute the bottom-left entry

    0×0+−1×1=−10\times 0+-1\times 1=-1

    Row two of B\mathbf{B} times column one of A\mathbf{A}.

  7. Compute the bottom-right entry

    0×−1+−1×0=00\times -1+-1\times 0=0

    Row two of B\mathbf{B} times column two of A\mathbf{A}.

  8. Check the determinant multiplies

    det⁡C=det⁡B×det⁡A=1\det \mathbf{C}=\det \mathbf{B}\times\det \mathbf{A}=1

    Determinants are multiplicative, which is a useful check.

  9. Find the images of the base vectors

    (01−10)(10)=(0−1),(01−10)(01)=(10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ -1 \end{pmatrix},\quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}

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

  10. Find the determinant

    det⁡M=1\det \mathbf{M}=1

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

  11. Compare with the standard forms

    (01−10)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

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

  12. 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.

  13. Test a second point

    (01−10)(11)=(1−1)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 \\ -1 \end{pmatrix}

    A second image confirms the transformation you have identified.

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

    (01−10)(20)=(0−2)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ -2 \end{pmatrix}

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

  15. Select the correct description of the composite

    Rotation of 270∘270^\circ anticlockwise about the origin

    This is the single transformation equivalent to PP then QQ.

Answer
Rotation of 270∘270^\circ anticlockwise about the origin
Question 3
8 markschallenging
The transformation PP is represented by A=(100−1)\mathbf{A}=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} and the transformation QQ is represented by B=(0110)\mathbf{B}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. Describe fully the single transformation equivalent to PP followed by QQ.
Show worked solution

Worked solution

  1. Identify the order of the product

    C=BA\mathbf{C}=\mathbf{B}\mathbf{A}

    The matrix of the first transformation goes on the right.

  2. Write down the product to be evaluated

    (0110)(100−1)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

    Multiply B\mathbf{B} on the left of A\mathbf{A}.

  3. Multiply the matrices

    (0110)(100−1)=(0−110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

    Use the row-by-column rule for each of the four entries.

  4. Compute the top-left entry

    0×1+1×0=00\times 1+1\times 0=0

    Row one of B\mathbf{B} times column one of A\mathbf{A}.

  5. Compute the top-right entry

    0×0+1×−1=−10\times 0+1\times -1=-1

    Row one of B\mathbf{B} times column two of A\mathbf{A}.

  6. Compute the bottom-left entry

    1×1+0×0=11\times 1+0\times 0=1

    Row two of B\mathbf{B} times column one of A\mathbf{A}.

  7. Compute the bottom-right entry

    1×0+0×−1=01\times 0+0\times -1=0

    Row two of B\mathbf{B} times column two of A\mathbf{A}.

  8. Check the determinant multiplies

    det⁡C=det⁡B×det⁡A=1\det \mathbf{C}=\det \mathbf{B}\times\det \mathbf{A}=1

    Determinants are multiplicative, which is a useful check.

  9. Find the images of the base vectors

    (0−110)(10)=(01),(0−110)(01)=(−10)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \end{pmatrix},\quad \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ 0 \end{pmatrix}

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

  10. Find the determinant

    det⁡M=1\det \mathbf{M}=1

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

  11. Compare with the standard forms

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

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

  12. 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.

  13. Test a second point

    (0−110)(11)=(−11)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ 1 \end{pmatrix}

    A second image confirms the transformation you have identified.

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

    (0−110)(20)=(02)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 2 \end{pmatrix}

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

  15. Select the correct description of the composite

    Rotation of 90∘90^\circ anticlockwise about the origin

    This is the single transformation equivalent to PP then QQ.

Answer
Rotation of 90∘90^\circ anticlockwise about the origin
Question 4
8 markschallenging
A linear transformation maps the point (1,0)(1,0) to (−12,32)(-\frac{1}{2},\frac{\sqrt{3}}{2}) and the point (0,1)(0,1) to (32,12)(\frac{\sqrt{3}}{2},\frac{1}{2}). Find the matrix that represents this transformation.
Show worked solution

Worked solution

  1. Recall that the columns are the images of the base vectors

    M=(MiMj)\mathbf{M}=\begin{pmatrix} \mathbf{M}\mathbf{i} & \mathbf{M}\mathbf{j} \end{pmatrix}

    The first column is the image of (1,0)(1,0), the second of (0,1)(0,1).

  2. Write the image of (1,0)(1,0) as the first column

    (−1232)\begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}

    This is given directly in the question.

  3. Write the image of (0,1)(0,1) as the second column

    (3212)\begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}

    This is the second piece of information given.

  4. Assemble the matrix

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

    Place the two image vectors side by side.

  5. Check the determinant

    det⁡M=−1\det \mathbf{M}=-1

    A quick check that the matrix is sensible.

  6. 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.

  7. 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.

  8. 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).

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. State the transformation matrix

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

    The columns are the images of the base vectors.

Answer
(−12323212)\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}
Question 5
8 markschallenging
The triangle ABCABC has vertices A(2,1), B(4,3), C(1,5)A(2,1),\ B(4,3),\ C(1,5). It is transformed by the matrix M=(0−110)\mathbf{M}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. Find the coordinates of the images A′A', B′B' and C′C'.
Show worked solution

Worked solution

  1. Write the vertices as a single matrix of columns

    (241135)\begin{pmatrix} 2 & 4 & 1 \\ 1 & 3 & 5 \end{pmatrix}

    Transforming all three vertices at once saves work.

  2. Find the image of AA

    (0−110)(21)=(−12)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    Multiply the matrix by the position vector of AA.

  3. Find the image of BB

    (0−110)(43)=(−34)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 4 \\ 3 \end{pmatrix}=\begin{pmatrix} -3 \\ 4 \end{pmatrix}

    Multiply the matrix by the position vector of BB.

  4. Find the image of CC

    (0−110)(15)=(−51)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 5 \end{pmatrix}=\begin{pmatrix} -5 \\ 1 \end{pmatrix}

    Multiply the matrix by the position vector of CC.

  5. Check the area scale factor

    ∣det⁡M∣=1\left|\det \mathbf{M}\right|=1

    The image triangle has this many times the area of the original.

  6. Note the origin is unchanged

    (0−110)(00)=(00)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}

    Linear transformations always fix the origin.

  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. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. State the three image points

    (−1, 2), (−3, 4), (−5, 1)(-1,\ 2),\ (-3,\ 4),\ (-5,\ 1)

    These are the vertices of the image triangle.

Answer
(−1, 2), (−3, 4), (−5, 1)(-1,\ 2),\ (-3,\ 4),\ (-5,\ 1)

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