Given the matrix A = i & 0 0 & −i .

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To show that A⁴ = I₂, we first need to calculate A² and then A⁴. Given the matrix A = i & 0 0 & −i . Recall that i² = −1. Step 1: Calculate A² A² = A × A; A² = i & 0 0 & −i × i & 0 0 & −i Perform matrix multiplication: A² = (i)(i) + (0)(0) & (i)(0) + (0)(−i) (0)(i) + (−i)(0) & (0)(0) + (−i)(−i) A² = i² + 0 & 0 + 0 0 + 0 & 0 + (−i)² Since i² = −1 and (−i)² = (−1)² i² = 1 · (−1) = −1: A² = −1 & 0 0 & −1 Step 2: Calculate A⁴ We can calculate A⁴ as A² × A². A⁴ = −1 & 0 0 & −1 × −1 & 0 0 & −1 Perform matrix multiplication: A⁴ = (−1)(−1) + (0)(0) & (−1)(0) + (0)(−1) (0)(−1) + (−1)(0) & (0)(0) + (−1)(−1) A⁴ = 1 + 0 & 0 + 0 0 + 0 & 0 + 1 A⁴ = 1 & 0 0 & 1 Step 3: Compare with I₂ The identity matrix of order 2 is I₂ = 1 & 0 0 & 1 . Since A⁴ = 1 & 0; 0 & 1 , we have A⁴ = I₂. Hence, it is proven that A⁴ = I₂.

Given the matrix A = i & 0 0 & -i .

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Calculate $A^2$

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