(a) The point P(3, -5) is rotated through an angle 60° anticlockwise about the origin. (i) Obtain the matrix for the rotation ; (ii) Find the image P' of th...

(a)(i) Rotation matrix through \(60^\circ\) anticlockwise about the origin.

\[R=\begin{pmatrix}\cos60^\circ&-\sin60^\circ\\\sin60^\circ&\cos60^\circ\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2}&-\tfrac{\sqrt3}{2}\\[2pt]\tfrac{\sqrt3}{2}&\tfrac{1}{2}\end{pmatrix}.\]

(ii) Image of \(P(3,-5).\)

\[P'=R\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}\tfrac12(3)-\tfrac{\sqrt3}{2}(-5)\\[2pt]\tfrac{\sqrt3}{2}(3)+\tfrac12(-5)\end{pmatrix}=\begin{pmatrix}\tfrac{3+5\sqrt3}{2}\\[2pt]\tfrac{3\sqrt3-5}{2}\end{pmatrix}.\]

Numerically \(P'\approx(5.83,\ 0.10).\)

(b) Transformation \(N:(x,y)\to(2x+3y,\ 3x-y).\)

(i) Matrix \(N=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.\)

(ii) Find a and b in \(N^2+aN+bI=0.\)

\[N^2=\begin{pmatrix}2&3\\3&-1\end{pmatrix}\begin{pmatrix}2&3\\3&-1\end{pmatrix}=\begin{pmatrix}13&3\\3&10\end{pmatrix}.\]

Then \(N^2+aN+bI=\begin{pmatrix}13+2a+b&3+3a\\3+3a&10-a+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.\)

Off-diagonal: \(3+3a=0\Rightarrow a=-1.\)

Top-left: \(13+2(-1)+b=0\Rightarrow b=-11.\)

Check bottom-right: \(10-(-1)+(-11)=0.\) Consistent.

\[a=-1,\qquad b=-11.\]

(This is the Cayley-Hamilton relation, since \(\text{trace}(N)=1\) and \(\det(N)=-11.\))