AI HL · AHL 5
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Composite transformations, area and invariant lines
AHL 5.9
Drag A and B. The columns of the matrix are the position vectors of A and B. Say what the determinant will be before you read it.
Select a point (Tab) and use the arrow keys to move it.
Drag v (orange). The teal arrow is Mv. Find the directions where Mv lies along the same line as v: those v are the eigenvectors.
Dragged close to an eigenvector line, v clicks onto it. Select v (Tab) and use the arrow keys for small steps.
\(A\) is a rotation of 90° anticlockwise about the origin. \(B\) is a reflection in the \(x\)-axis. Find the matrix for \(A\) followed by \(B\), and describe it.
Have a go first. Then open the working one line at a time.
\(A = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},\quad B = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\)
From the formula booklet, or from where \((1, 0)\) and \((0, 1)\) go.
\(BA = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\)
The first transformation goes on the right.
A reflection in the line \(y = -x\)
\((1, 0)\) goes to \((0, -1)\) and \((0, 1)\) goes to \((-1, 0)\).
\(AB = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\)
\(B\) followed by \(A\) is a reflection in \(y = x\). Different.
The point \((3, 1)\) is enlarged with scale factor 2 about the origin and then translated by \(\begin{pmatrix}-4 \\ 5\end{pmatrix}\). Find the image. Then find the image if the translation is done first.
Have a go first. Then open the working one line at a time.
\(2\begin{pmatrix}3 \\ 1\end{pmatrix} + \begin{pmatrix}-4 \\ 5\end{pmatrix} = \begin{pmatrix}2 \\ 7\end{pmatrix}\)
\((2, 7)\).
\(2\left(\begin{pmatrix}3 \\ 1\end{pmatrix} + \begin{pmatrix}-4 \\ 5\end{pmatrix}\right) = \begin{pmatrix}-2 \\ 12\end{pmatrix}\)
\((-2, 12)\). The order changes the answer.
A point \(P\) is transformed by \(M = \begin{pmatrix}3 & 1 \\ 2 & 2\end{pmatrix}\) and then translated by \(\begin{pmatrix}-4 \\ 1\end{pmatrix}\). Its image is \((5, 11)\). Find \(P\).
Have a go first. Then open the working one line at a time.
\(M\begin{pmatrix}x \\ y\end{pmatrix} + \begin{pmatrix}-4 \\ 1\end{pmatrix} = \begin{pmatrix}5 \\ 11\end{pmatrix}\)
The matrix first, then the translation is added.
\(M\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}5 \\ 11\end{pmatrix} - \begin{pmatrix}-4 \\ 1\end{pmatrix} = \begin{pmatrix}9 \\ 10\end{pmatrix}\)
Going back, undo the last step first.
\(\begin{pmatrix}x \\ y\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2 & -1 \\ -2 & 3\end{pmatrix}\begin{pmatrix}9 \\ 10\end{pmatrix} = \begin{pmatrix}2 \\ 3\end{pmatrix}\)
\(\det M = 4\). \(P\) is \((2, 3)\).
\(\begin{pmatrix}3 & 1 \\ 2 & 2\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix} + \begin{pmatrix}-4 \\ 1\end{pmatrix} = \begin{pmatrix}5 \\ 11\end{pmatrix}\)
Forwards again gives the image.
The triangle with vertices \((0, 0)\), \((4, 0)\) and \((1, 3)\) is transformed by \(\begin{pmatrix}3 & 1 \\ -2 & 2\end{pmatrix}\). Find the area of the image.
Have a go first. Then open the working one line at a time.
\(\frac{1}{2}(4)(3) = 6\)
Base 4, height 3.
\(3(2) - 1(-2) = 8\)
The scale factor for area.
\(8 \times 6 = 48\)
No need to find the image points.
The unit square is transformed by \(\begin{pmatrix}1 & 4 \\ 2 & 3\end{pmatrix}\). Find the area of the image.
Have a go first. Then open the working one line at a time.
\(1(3) - 4(2) = -5\)
Leading diagonal, less the other diagonal.
\(|-5| \times 1 = 5\)
Use the size of the determinant. The negative sign means the square has been turned over.
\(M = \begin{pmatrix}3 & 1 \\ 1 & 1\end{pmatrix}\). Find the area of the image of the unit square under \(M^5\).
Have a go first. Then open the working one line at a time.
\(3(1) - 1(1) = 2\)
Each use of \(M\) doubles the area.
\(2^5 = 32\)
No need to find \(M^5\).
Find the equations of the invariant lines through the origin for the transformation with matrix \(\begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}1 \\ 1\end{pmatrix}\) for \(\lambda = 5\) and \(\begin{pmatrix}1 \\ -2\end{pmatrix}\) for \(\lambda = 2\)
Found in the eigenvalues section.
\(y = x\) and \(y = -2x\)
The gradient is the bottom number over the top number.
Points on \(y = x\) move 5 times as far from the origin. Points on \(y = -2x\) move twice as far.
They stay on their line.
The transformation \(T\) has matrix \(\begin{pmatrix}3 & -2 \\ 1 & 0\end{pmatrix}\). Show that every point on the line \(y = x\) is invariant, and find the other invariant line through the origin.
Have a go first. Then open the working one line at a time.
\(M\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}x \\ y\end{pmatrix}\)
This is the eigenvalue \(\lambda = 1\).
\(M\begin{pmatrix}t \\ t\end{pmatrix} = \begin{pmatrix}3t - 2t \\ t\end{pmatrix} = \begin{pmatrix}t \\ t\end{pmatrix}\)
Every point \((t, t)\) stays where it is. This is a line of invariant points.
\(\lambda^2 - 3\lambda + 2 = 0,\quad \lambda = 1 \text{ or } 2\)
The other eigenvalue is 2.
\(x - 2y = 0\)
So \(y = \tfrac{1}{2}x\). Points on this line stay on it but move twice as far from the origin: it is an invariant line, not a line of invariant points.
\(T\) is a horizontal stretch with scale factor 2, then a reflection in the line \(y = x\), then a rotation of 90° anticlockwise about the origin. Find the single matrix for \(T\), and the area of the image of a shape of area 5.
Have a go first. Then open the working one line at a time.
\(S = \begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix},\quad F = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\quad R = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\)
In the order done: stretch, reflection, rotation.
\(RFS = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}\)
The first transformation is closest to the point, so it is on the right.
\(\begin{pmatrix}-2 & 0 \\ 0 & 1\end{pmatrix}\)
Work from the right: \(FS = \begin{pmatrix}0 & 1 \\ 2 & 0\end{pmatrix}\) first.
\(\det = -2,\quad \text{area scale factor} = |{-2}| = 2\)
The negative sign says the shape is flipped over. The area is 5 × 2 = 10.