IB Mathematics AHL 1.14 Definition of a matrix AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Markscheme
a) Compute \( R(\alpha) \times R(\alpha) \), where \( R(\alpha) = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \):
Compute \( R(\alpha) \times R(\alpha) = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \times \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \) [M1].
Calculate each element:
– Top-left (row 1, column 1): \( \cos(\alpha) \times \cos(\alpha) + (-\sin(\alpha)) \times \sin(\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \).
– Top-right (row 1, column 2): \( \cos(\alpha) \times (-\sin(\alpha)) + (-\sin(\alpha)) \times \cos(\alpha) = -\cos(\alpha) \sin(\alpha) – \sin(\alpha) \cos(\alpha) = -2 \times \sin(\alpha) \times \cos(\alpha) \).
– Bottom-left (row 2, column 1): \( \sin(\alpha) \times \cos(\alpha) + \cos(\alpha) \times \sin(\alpha) = 2 \times \sin(\alpha) \times \cos(\alpha) \).
– Bottom-right (row 2, column 2): \( \sin(\alpha) \times (-\sin(\alpha)) + \cos(\alpha) \times \cos(\alpha) = -\sin^2(\alpha) + \cos^2(\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \).
Result: \( \begin{pmatrix} \cos^2(\alpha) – \sin^2(\alpha) & -2 \times \sin(\alpha) \times \cos(\alpha) \\ 2 \times \sin(\alpha) \times \cos(\alpha) & \cos^2(\alpha) – \sin^2(\alpha) \end{pmatrix} \) [A1].
Final Answer: \( R(\alpha) \times R(\alpha) = \begin{pmatrix} \cos^2(\alpha) – \sin^2(\alpha) & -2 \times \sin(\alpha) \times \cos(\alpha) \\ 2 \times \sin(\alpha) \times \cos(\alpha) & \cos^2(\alpha) – \sin^2(\alpha) \end{pmatrix} \). [2]
b) (i) Compare the elements in the first row, second column:
From given: \( R(2\alpha) = \begin{pmatrix} \cos(2\alpha) & -\sin(2\alpha) \\ \sin(2\alpha) & \cos(2\alpha) \end{pmatrix} \).
From part (a): \( R(\alpha) \times R(\alpha) = \begin{pmatrix} \cos^2(\alpha) – \sin^2(\alpha) & -2 \times \sin(\alpha) \times \cos(\alpha) \\ 2 \times \sin(\alpha) \times \cos(\alpha) & \cos^2(\alpha) – \sin^2(\alpha) \end{pmatrix} \).
First row, second column: in \( R(2\alpha) \), \( -\sin(2\alpha) \); in \( R(\alpha) \times R(\alpha) \), \( -2 \times \sin(\alpha) \times \cos(\alpha) \).
Equate: \( -\sin(2\alpha) = -2 \times \sin(\alpha) \times \cos(\alpha) \).
Multiply by \(-1\): \( \sin(2\alpha) = 2 \times \sin(\alpha) \times \cos(\alpha) \) [A1].
Final Answer: \( \sin(2\alpha) = 2 \times \sin(\alpha) \times \cos(\alpha) \). [1]
b) (ii) Compare the elements in the first row, first column, and rewrite using \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \):
First row, first column: in \( R(2\alpha) \), \( \cos(2\alpha) \); in \( R(\alpha) \times R(\alpha) \), \( \cos^2(\alpha) – \sin^2(\alpha) \).
Equate: \( \cos(2\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \) [A1].
Use \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \): \( \cos^2(\alpha) = 1 – \sin^2(\alpha) \).
Substitute: \( \cos(2\alpha) = \cos^2(\alpha) – \sin^2(\alpha) = (1 – \sin^2(\alpha)) – \sin^2(\alpha) = 1 – 2 \times \sin^2(\alpha) \) [A1].
Final Answer: \( \cos(2\alpha) = 1 – 2 \times \sin^2(\alpha) \). [2]
Total: [5 marks]