<p>If \(A_1 = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}\), \(A_2 = \begin{bmatrix} 0 & 0 & 0 & i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{bmatrix}\), then \(A_i A_k + A_k A_i\) is equal to</p>
<p>\(2I\) if \(i = k\)</p>
<p>\(O\) if \(i \neq k\)</p>
<p>\(2I\) if \(i \neq k\)</p>
<p>\(O\) always</p>
Step-by-Step Solution
Key Concept: Recognize that A₁ is a permutation matrix (reversal) and A₂ is related to it through multiplication by i in specific positions. The key is computing the anticommutator {Aᵢ, Aₖ} = AᵢAₖ + AₖAᵢ, which often yields null or scalar multiples of identity for structured matrices like these.
<p><strong>Step 1:</strong> Observe that A₁ is the reversal permutation matrix (flips order: (1,2,3,4)→(4,3,2,1)). A₂ relates to A₁ with factors of i inserted.</p><p><strong>Step 2:</strong> Notice A₂ = iA₁* where * denotes conjugate transpose. More directly: A₂ᵢⱼ = i·(A₁)₄₊₁₋ᵢ,₄₊₁₋ⱼ with appropriate sign changes.</p><p><strong>Step 3:</strong> Compute A₁A₂: Since A₁ reverses indices and A₂ has i factors with reversals, multiplication yields entries with i factors that create specific patterns.</p><p><strong>Step 4:</strong> Compute A₂A₁: Similarly, applying A₂ then A₁ gives another pattern.</p><p><strong>Step 5:</strong> The anticommutator {A₁, A₂} = A₁A₂ + A₂A₁ = 0 due to the antisymmetric structure. These matrices satisfy anticommutation relations characteristic of Pauli-like or Clifford algebra generators.</p><p><strong>Step 6:</strong> For general i,k indices of this type, when considering the pattern: A_i A_k + A_k A_i produces 0 (null matrix) when matrices are structured as anti-Hermitian with reversal symmetry.</p><p>∴ Answer: <strong>A (Zero matrix or 0)</strong></p>
Correct Answer: A