Matrices & Determinants
Determinants
Grade 12
Question:
<p>Let \(\alpha = \dfrac{\pi}{5}\) and \(A = \begin{bmatrix} \cos\dfrac{\pi}{5} & \sin\dfrac{\pi}{5} \\ -\sin\dfrac{\pi}{5} & \cos\dfrac{\pi}{5} \end{bmatrix}\), then find \(\det(A + A^2 + A^3 + A^4)\)</p>
<p>(a) \(4\cos^2\dfrac{\pi}{5}\cos^2\dfrac{\pi}{10}\)</p>
<p>(b) \(-\cos^2\dfrac{\pi}{5}\cos^2\dfrac{\pi}{10}\)</p>
<p>(c) \(16\cos^2\dfrac{\pi}{5}\cos^2\dfrac{\pi}{10}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that A is a rotation matrix with det(A) = 1, and the sum A + A² + A³ + A⁴ forms a geometric series that can be factored using matrix algebra to yield a determinant of 0.
<p><strong>Step 1:</strong> Recognize that A is a rotation matrix by angle π/5. Calculate: det(A) = cos²(π/5) + sin²(π/5) = 1</p><p><strong>Step 2:</strong> Note that A⁵ = I (five rotations of π/5 equal 2π), so A satisfies A⁵ - I = 0</p><p><strong>Step 3:</strong> Factor the geometric series: A + A² + A³ + A⁴ = A(I + A + A² + A³) = A · (A⁵ - I)/(A - I) = A · 0/(A - I)</p><p><strong>Step 4:</strong> Alternatively, note that (A - I)(A + A² + A³ + A⁴) = A⁵ - A = I - A, so the matrix (A + A² + A³ + A⁴) is singular when this system is analyzed.</p><p><strong>Step 5:</strong> Direct verification: A + A² + A³ + A⁴ has rank < 4, making it singular.</p><p>∴ det(A + A² + A³ + A⁴) = <strong>0</strong></p>
Correct Answer: C