(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 the point P under the rotation.
(b) A linear transformation is given by (N : (x, y) to (2x + 3y, 3x – y)).
(i) Write down the matrix N of the transformation ; (ii) If (N^{2} + aN + bI = 0), where a, b (in) R, (I) is the (2 times 2) matrix and (0) is the (2 times 2) null matrix, find the values of a and b.
Explanation
(a)(i) The matrix of rotation of point anticlockwise through angle 60° about the origin is given by:
(M = begin{pmatrix} cos theta & -sin theta \ sin theta & cos theta end{pmatrix} = begin{pmatrix} cos 60° & -sin 60° \ sin 60° & cos 60° end{pmatrix})
= (begin{pmatrix} frac{1}{2} & frac{-sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix})
(ii) Using the transformation,
(T 😛 to MP) or (T : begin{pmatrix} x \ y end{pmatrix} to begin{pmatrix} cos theta & -sin theta \ sin theta & cos theta end{pmatrix} begin{pmatrix} x \ y end{pmatrix})
(T : begin{pmatrix} 3 \ 5 end{pmatrix} to begin{pmatrix} frac{1}{2} & frac{-sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix} begin{pmatrix} 3 \ 5 end{pmatrix})
Image of P, P’ = (begin{pmatrix} frac{3 + 5sqrt{3}}{2} \ frac{3sqrt{3} – 5}{2} end{pmatrix})
(b) (T : begin{pmatrix} x \ y end{pmatrix} to begin{pmatrix} 2x + 3y \ 3x – y end{pmatrix} = begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} begin{pmatrix} x \ y end{pmatrix})
(i) (N = begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix})
(ii) (N^{2} + aN + bI = 0)
(begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} ^{2} + a begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} + b begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} = begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix})
(begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} + a begin{pmatrix} 2 & 3 \ 3 & -1 end{pmatrix} + b begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} = begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix})
(begin{pmatrix} 4 + 9 & 6 – 3 \ 6 – 3 & 9 + 1 end{pmatrix} + begin{pmatrix} 2a & 3a \ 3a & -a end{pmatrix} + begin{pmatrix} b & 0 \ 0 & b end{pmatrix} = begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix})
Adding corresponding elements of each matrix,
(13 + 2a + b = 0 implies 2a + b = -13 … (1))
(3 + 3a = 0 implies 3a = -3 ; a = -1)
(2(-1) + b = -13 implies -2 + b = -13)
(b = -13 + 2 = -11)
(therefore a = -1 ; b = -11)