<p>If \(A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}\), \(n \in \mathbb{N}\), then trace of \(A^{4n}\) equals to ___________.</p>
Step-by-Step Solution
Key Concept: Recognize that A is a scalar multiple of the identity matrix (A = iI), so A^(4n) = i^(4n)·I, and use the periodicity of powers of i with period 4.
<p><strong>Step 1:</strong> Observe that the matrix A can be written as:</p><p>A = iI where I is the 2×2 identity matrix</p><p><strong>Step 2:</strong> Use the property that (kI)^n = k^n·I for scalar k and identity matrix I:</p><p>A^(4n) = (iI)^(4n) = i^(4n)·I</p><p><strong>Step 3:</strong> Simplify i^(4n) using the periodicity of powers of i:</p><p>i¹ = i, i² = -1, i³ = -i, i⁴ = 1, i⁵ = i, ...</p><p>Since 4n is divisible by 4: i^(4n) = (i⁴)^n = 1^n = 1</p><p><strong>Step 4:</strong> Therefore:</p><p>A^(4n) = 1·I = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}</p><p><strong>Step 5:</strong> The trace of a matrix is the sum of diagonal elements:</p><p>Trace(A^(4n)) = 1 + 1 = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2