Answer: (a) Shear with the \(x\)-axis fixed; \((0,1)\mapsto(\frac32,1)\).

(b) Shear with the \(y\)-axis fixed; \((1,0)\mapsto(1,\frac32)\).

(c) \(\det(AB)=1\), so the area scale factor is \(1\).

(d) \(y=-2x\) and \(y=\frac12x\).

(a)

Under \(A\),

\(\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x+\frac32y\\y\end{pmatrix}.\)

All points on the \(x\)-axis have \(y=0\), so they are unchanged. The point \((0,1)\) maps to \((\frac32,1)\). Therefore the transformation is a shear with the \(x\)-axis fixed.

(b)

Under \(B\),

\(\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x\\\frac32x+y\end{pmatrix}.\)

All points on the \(y\)-axis have \(x=0\), so they are unchanged. The point \((1,0)\) maps to \((1,\frac32)\). Therefore the transformation is a shear with the \(y\)-axis fixed.

(c)

Compute

\(AB=\begin{pmatrix}1&\frac32\\0&1\end{pmatrix}\begin{pmatrix}1&0\\\frac32&1\end{pmatrix}=\begin{pmatrix}\frac{13}{4}&\frac32\\\frac32&1\end{pmatrix}.\)

Then

\(\det(AB)=\frac{13}{4}\cdot1-\frac32\cdot\frac32=\frac{13}{4}-\frac94=1.\)

The determinant gives the area scale factor, so the two triangles have the same area.

(d)

Using

\(AB=\begin{pmatrix}\frac{13}{4}&\frac32\\\frac32&1\end{pmatrix},\)

a point \((x,y)\) maps to

\((X,Y)=\left(\frac{13}{4}x+\frac32y,\frac32x+y\right).\)

An invariant line through the origin has the form \(y=mx\). Substitute \(y=mx\) and require \(Y=mX\):

\(\frac32x+mx=m\left(\frac{13}{4}x+\frac32mx\right).\)

For \(x\ne0\),

\(\frac32+m=\frac{13}{4}m+\frac32m^2.\)

Hence

\(m^2+\frac32m-1=0=(m+2)\left(m-\frac12\right).\)

So \(m=-2\) or \(m=\frac12\), giving \(y=-2x\) and \(y=\frac12x\).

Mark scheme