P and Q are two linear transformations in the X-Y plane defined by P: (x, y) → (-3x + 6y, 4x + y) and Q: (x, y) → (2x-3y, -4x - 6y). (a) Write down the matr...

Answer Details

(a) Reading the coefficients of \(x\) and \(y\) from each mapping:

\[P=\begin{pmatrix}-3&6\\4&1\end{pmatrix},\qquad Q=\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}\]

(b) Image of \((-2,-3)\) under \(Q\): \((2(-2)-3(-3),\ -4(-2)-6(-3))=(-4+9,\ 8+18)=(5,26)\).

(c) "Q followed by P" means \(P\) acts second, so the single matrix is \(PQ\):

\[PQ=\begin{pmatrix}-3&6\\4&1\end{pmatrix}\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}=\begin{pmatrix}-30&-27\\4&-18\end{pmatrix}\]

i.e. \((x,y)\to(-30x-27y,\ 4x-18y)\).

(d) Image of \((1,4)\): \((-30(1)-27(4),\ 4(1)-18(4))=(-30-108,\ 4-72)=(-138,-68)\).

(e) Anticlockwise rotation of \(225^{\circ}\): \(\begin{pmatrix}\cos225^{\circ}&-\sin225^{\circ}\\\sin225^{\circ}&\cos225^{\circ}\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{pmatrix}\).

Apply to \((-\sqrt2,\ 2\sqrt2)\):

\(x'=-\tfrac{\sqrt2}{2}(-\sqrt2)+\tfrac{\sqrt2}{2}(2\sqrt2)=1+2=3\)

\(y'=-\tfrac{\sqrt2}{2}(-\sqrt2)-\tfrac{\sqrt2}{2}(2\sqrt2)=1-2=-1\)

So \(P'=(3,-1)\).