Composite transformations, area and invariant lines Problem set A Answers

1. (a) \(\begin{pmatrix}5 & 4 \\ 1 & 2\end{pmatrix}\)
2. (b) \(\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}\)
3. (c) \(\begin{pmatrix}0 & -2 \\ 2 & 0\end{pmatrix}\)

1. (a) 35
2. (b) 36
3. (c) 8

1. (a) \((1, 9)\)
2. (b) \((-3, -3)\)

1. (a) \(y = x\) and \(y = -\frac{1}{2}x\)
2. (b) \(y = 2x\) and \(y = -x\)

1. (a) \(\lambda = 1 \text{ or } 3\)
2. (b) \(y = 0\)
3. (c) \(y = x\); points move 3 times as far from the origin.

\(\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\): a reflection in the \(x\)-axis.

\(\det M = 3\), so the area is \(3^4 = 81\).

A 90° turn changes the direction of every vector, so no line maps to itself (no real eigenvalues). A 180° turn sends every vector to its opposite, which is on the same line: every line through the origin is invariant (eigenvalue −1).

1. (a) \(\begin{pmatrix}0 & -3 \\ -3 & 0\end{pmatrix}\)
2. (b) 36

(a) \(\begin{pmatrix}0 & -0.5 \\ 0.5 & 0\end{pmatrix}\). (b) 8 and 2 (\(\det T = 0.25\)). (c) A geometric series: \(\dfrac{8}{1 - 0.25} = \dfrac{32}{3}\).