Composite transformations, area and invariant lines Problem set B Worked solutions

Find the single matrix for each pair of transformations.

1. (a) \(\begin{pmatrix}2 & -1 \\ 0 & 3\end{pmatrix}\) followed by \(\begin{pmatrix}1 & 0 \\ 4 & 1\end{pmatrix}\)

   • The matrix of the second transformation goes on the left: \(\begin{pmatrix}1 & 0 \\ 4 & 1\end{pmatrix}\begin{pmatrix}2 & -1 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}2 & -1 \\ 8 & -1\end{pmatrix}\).

2. (b) Reflection in the \(x\)-axis, followed by rotation 90° anticlockwise about the origin

   • Reflection in the \(x\)-axis: \(\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\). Rotation 90° anticlockwise: \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\).
   • \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\).

3. (c) Enlargement with scale factor 3, followed by rotation 90° clockwise

   • Enlargement: \(\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}\). Rotation 90° clockwise: \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\).
   • \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}0 & 3 \\ -3 & 0\end{pmatrix}\).

Find the area of the image.

1. (a) The unit square under \(\begin{pmatrix}5 & -2 \\ 3 & 4\end{pmatrix}\)

   • Area scale factor = \(|\det| = |(5)(4) - (-2)(3)| = 26\). The unit square has area 1, so the image has area 26.

2. (b) The triangle with vertices \((0,0),\ (4,0),\ (0,3)\) under \(\begin{pmatrix}3 & -1 \\ 2 & 2\end{pmatrix}\)

   • The triangle has area \(\tfrac{1}{2}(4)(3) = 6\). Area scale factor = \(|\det| = |(3)(2) - (-1)(2)| = 8\).
   • Image area = \(6 \times 8 = 48\).

3. (c) A rectangle of area 8 under \(\begin{pmatrix}2.5 & 0 \\ 0 & 1\end{pmatrix}\)

   • Area scale factor = \(|\det| = (2.5)(1) - 0 = 2.5\).
   • Image area = \(8 \times 2.5 = 20\).

Find the image of each point.

1. (a) \((3, -2)\): enlargement, scale factor 4, then translation by \(\begin{pmatrix}-6 \\ 3\end{pmatrix}\)

   • Enlargement: \((3,-2) \to (12,-8)\).
   • Translation: \((12 - 6,\ -8 + 3) = (6, -5)\).

2. (b) \((-1, 5)\): translation by \(\begin{pmatrix}4 \\ -3\end{pmatrix}\), then rotation 90° clockwise about the origin

   • Translation: \((-1 + 4,\ 5 - 3) = (3, 2)\).
   • Rotation 90° clockwise: \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}3 \\ 2\end{pmatrix} = \begin{pmatrix}2 \\ -3\end{pmatrix}\), so \((2, -3)\).

Find the equations of the invariant lines through the origin for each matrix.

1. (a) \(\begin{pmatrix}6 & 1 \\ 2 & 5\end{pmatrix}\)

   • \(\det(M - \lambda I) = (6 - \lambda)(5 - \lambda) - (1)(2) = 0\)
   • \(\lambda^2 - 11\lambda + 28 = 0\), so \((\lambda - 7)(\lambda - 4) = 0\): \(\lambda = 7\) or \(\lambda = 4\).
   • For \(\lambda = 7\): \((M - 7I)v = 0\) gives \(-x + y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
   • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(2x + y = 0\), so \(v = \begin{pmatrix}1 \\ -2\end{pmatrix}\).
   • Invariant lines through the origin have the direction of an eigenvector: \(y = x\) and \(y = -2x\).

2. (b) \(\begin{pmatrix}7 & -3 \\ 2 & 2\end{pmatrix}\)

   • \(\det(M - \lambda I) = (7 - \lambda)(2 - \lambda) - (-3)(2) = 0\)
   • \(\lambda^2 - 9\lambda + 20 = 0\), so \((\lambda - 5)(\lambda - 4) = 0\): \(\lambda = 5\) or \(\lambda = 4\).
   • For \(\lambda = 5\): \((M - 5I)v = 0\) gives \(2x - 3y = 0\), so \(v = \begin{pmatrix}3 \\ 2\end{pmatrix}\).
   • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(3x - 3y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
   • Invariant lines through the origin have the direction of an eigenvector: \(y = \frac{2}{3}x\) and \(y = x\).

The transformation \(T\) has matrix \(\begin{pmatrix}1 & 0 \\ 3 & 4\end{pmatrix}\).

1. (a) Find the eigenvalues of the matrix.

   • \(\lambda^2 - 5\lambda + 4 = 0\)
   • \(\lambda = 1 \text{ or } 4\)

2. (b) Find the equation of the line of invariant points.

   • Invariant points satisfy \(M\mathbf{v} = \mathbf{v}\), the eigenvalue 1.
   • Eigenvector: \(\begin{pmatrix}-1 \\ 1\end{pmatrix}\), so the line is \(y = -x\).

3. (c) Find the equation of the other invariant line through the origin, and say how far points on it move from the origin.

   • Eigenvector for \(4\): \(\begin{pmatrix}0 \\ 1\end{pmatrix}\), so the line is \(x = 0\).
   • The eigenvalue is 4, so each point on it moves 4 times as far from the origin, along the line.

Describe the single transformation equal to a reflection in the \(y\)-axis followed by a rotation of 90° anticlockwise about the origin.

• Reflection in the \(y\)-axis: \(\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}\). Rotation 90° anticlockwise: \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\).
• The second transformation goes on the left: \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\).
• This matrix swaps \(x\) and \(y\) and changes both signs, which is a reflection in the line \(y = -x\).

\(M = \begin{pmatrix}3 & 2 \\ 2 & 3\end{pmatrix}\). Find the area of the image of the unit square under \(M^3\).

• \(\det M = (3)(3) - (2)(2) = 5\).
• \(\det(M^3) = (\det M)^3 = 5^3 = 125\).
• The unit square has area 1, so the image has area 125.

Explain why a rotation about the origin through 120° has no invariant lines through the origin, and an enlargement with scale factor 3, centre the origin, has infinitely many.

• An invariant line through the origin is the direction of an eigenvector, so \(Mv = \lambda v\) for some real \(\lambda\).
• A 120° rotation turns every vector through 120°, so no vector stays on its own line and there are no real eigenvalues.
• The enlargement has \(M = 3I\), so \(Mv = 3v\) for every vector: every line through the origin is invariant, with eigenvalue 3.

Transformation \(T\) is a vertical stretch with scale factor 3, then a rotation of 90° anticlockwise about the origin, then a reflection in the \(y\)-axis.

1. (a) Find the single matrix for \(T\).

   • Matrices in the order done: \(\begin{pmatrix}1 & 0 \\ 0 & 3\end{pmatrix}\), \(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\), \(\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}\).
   • The first is on the right: \(M = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}1 & 0 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}0 & 3 \\ 1 & 0\end{pmatrix}\)

2. (b) A shape has area 6. Find the area of its image under \(T\).

   • \(\det M = -3\)
   • Area scale factor \(|\det M| = 3\), so the area is \(3 \times 6 = 18\)

A logo is made from a triangle with vertices \((0,0),\ (6,0)\) and \((0,4)\), and its images under \(T,\ T^2,\ T^3,\ \ldots\), where \(T\) is an enlargement with scale factor \(\frac{1}{3}\) followed by a rotation of 90° clockwise about the origin. (a) Find the matrix of \(T\). (b) Find the area of the first triangle and of its image under \(T\). (c) Find the total area of all the triangles.

• (a) Enlargement: \(\begin{pmatrix}\frac{1}{3} & 0 \\ 0 & \frac{1}{3}\end{pmatrix}\). Rotation 90° clockwise: \(\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\). \(T = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}\frac{1}{3} & 0 \\ 0 & \frac{1}{3}\end{pmatrix} = \begin{pmatrix}0 & \frac{1}{3} \\ -\frac{1}{3} & 0\end{pmatrix}\).
• (b) First triangle: \(\tfrac{1}{2}(6)(4) = 12\). \(\det T = (0)(0) - (\frac{1}{3})(-\frac{1}{3}) = \frac{1}{9}\), so the image has area \(12 \times \frac{1}{9} = \frac{4}{3}\).
• (c) The areas are \(12,\ \frac{4}{3},\ \frac{4}{27},\ \ldots\), a geometric series with first term 12 and ratio \(\frac{1}{9}\).
• \(S_\infty = \dfrac{12}{1 - \frac{1}{9}} = \dfrac{27}{2}\).