(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.
(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.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
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,
\(T :P \to MP\) 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' = \(\begin{pmatrix} \frac{3 + 5\sqrt{3}}{2} \\ \frac{3\sqrt{3} - 5}{2} \end{pmatrix}\)
(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,
\(13 + 2a + b = 0 \implies 2a + b = -13 ... (1)\)
\(3 + 3a = 0 \implies 3a = -3 ; a = -1\)
\(2(-1) + b = -13 \implies -2 + b = -13\)
\(b = -13 + 2 = -11\)
\(\therefore a = -1 ; b = -11\)
(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,
\(T :P \to MP\) 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' = \(\begin{pmatrix} \frac{3 + 5\sqrt{3}}{2} \\ \frac{3\sqrt{3} - 5}{2} \end{pmatrix}\)
(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,
\(13 + 2a + b = 0 \implies 2a + b = -13 ... (1)\)
\(3 + 3a = 0 \implies 3a = -3 ; a = -1\)
\(2(-1) + b = -13 \implies -2 + b = -13\)
\(b = -13 + 2 = -11\)
\(\therefore a = -1 ; b = -11\)