<p>If \(A\) and \(B\) are square matrices such that \(A^{2006} = O\) and \(AB = A + B\), then \(\det(B)\) equals</p>
Step-by-Step Solution
Key Concept: From AB = A + B, rearrange to A(B - I) = B, then use the nilpotency condition A^2006 = O to show that B must be singular. The equation AB - A - B = 0 can be rewritten as A(B - I) - B = 0, leading to the conclusion that det(B) = 0.
<p><strong>Step 1:</strong> Start with the given equation: AB = A + B</p><p><strong>Step 2:</strong> Rearrange: AB - A - B = 0, which gives A(B - I) = B</p><p><strong>Step 3:</strong> Since A^2006 = O (nilpotent), we know det(A) = 0 and A is not invertible.</p><p><strong>Step 4:</strong> From A(B - I) = B, apply determinant to both sides: det(A)·det(B - I) = det(B)</p><p><strong>Step 5:</strong> Since det(A) = 0, the left side equals 0, so det(B) = 0.</p><p><strong>Verification:</strong> Rewrite as A(B - I) - B = 0 ⟹ (A - I)B = A. Since A is nilpotent with A^2006 = O, repeated multiplication shows B must be singular.</p><p>∴ <strong>det(B) = 0</strong></p>
Correct Answer: B