1. **Problem:** Given matrix \( \Lambda = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} \), find the characteristic equation if \( \Lambda - 2I \) is singular. 2. Since \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then $$ \Lambda - 2I = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1-2 & 3 \\ 2 & 2-2 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix} $$ 3. For \( \Lambda - 2I \) to be singular, its determinant must be zero: $$ \det(\Lambda - 2I) = (-1)(0) - (3)(2) = -6 \neq 0 $$ This means the matrix \( \Lambda - 2I \) is not singular, so \( \lambda = 2 \) is not an eigenvalue. Instead, find the characteristic polynomial: $$ \det(\Lambda - \lambda I) = 0 $$ $$ \det \begin{bmatrix} 1-\lambda & 3 \\ 2 & 2-\lambda \end{bmatrix} = (1-\lambda)(2-\lambda) - 6 = 0 $$ 4. Expand and simplify: $$ (1-\lambda)(2-\lambda) - 6 = (2 - \lambda - 2\lambda + \lambda^{2}) - 6 = \lambda^{2} - 3\lambda - 4 = 0 $$ 5. **Answer for the first part:** The characteristic equation is $$ \boxed{\lambda^{2} - 3\lambda - 4 = 0} $$ --- **Q2. i) Convert \(\frac{1}{4}^\circ\) to radians:** $$ \frac{1}{4}^\circ = \frac{1}{4} \times \frac{\pi}{180} = \frac{\pi}{720} \text{ radians} $$ ii) Find \( \sin \left( \frac{41\pi}{4} \right) \): - The sine function has period \( 2\pi \). Reduce angle modulo \( 2\pi \): $$ \frac{41\pi}{4} \mod 2\pi = \frac{41\pi}{4} - 10\pi = \frac{41\pi}{4} - \frac{40\pi}{4} = \frac{\pi}{4} $$ - So, $$ \sin \left( \frac{41\pi}{4} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$ iii) Determine the quadrant for \( \theta \) with \( \tan \theta < 0 \) and \( \sec \theta > 0 \): - \( \sec \theta = \frac{1}{\cos \theta} > 0 \) means \( \cos \theta > 0 \). - \( \tan \theta = \frac{\sin \theta}{\cos \theta} < 0 \) means \( \sin \theta \) and \( \cos \theta \) have opposite signs. - Since \( \cos \theta > 0 \), \( \sin \theta < 0 \). - This is true in **Quadrant IV**. iv) Express \( 2 \cos 35^\circ \cos 75^\circ \) as a sum or difference: Use the product-to-sum formula: $$ 2 \cos A \cos B = \cos(A+B) + \cos(A-B) $$ So, $$ 2 \cos 35^\circ \cos 75^\circ = \cos(35^\circ + 75^\circ) + \cos(35^\circ - 75^\circ) = \cos 110^\circ + \cos(-40^\circ) $$ Since \( \cos(-x) = \cos x \), $$ = \cos 110^\circ + \cos 40^\circ $$ **Summary answers:** - Characteristic equation: \( \lambda^{2} - 3\lambda - 4 = 0 \) (option b) - \( \frac{1}{4}^\circ = \frac{\pi}{720} \) radians - \( \sin \left( \frac{41\pi}{4} \right) = \frac{\sqrt{2}}{2} \) - Quadrant IV for given conditions - \( 2 \cos 35^\circ \cos 75^\circ = \cos 110^\circ + \cos 40^\circ \)