<p>If \(A\) and \(B\) are two nonzero square matrices of the same order such that the product \(AB = O\), then</p>
<p>both \(A\) and \(B\) must be singular</p>
<p>exactly one of them must be singular</p>
<p>both of them are nonsingular</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: When AB = O for nonzero matrices, both A and B must be singular (non-invertible) with rank(A) + rank(B) ≤ n. The null space of B must contain the column space of A.
<p><strong>Step 1:</strong> Assume A is invertible. Then AB = O implies A⁻¹(AB) = A⁻¹O, giving B = O. This contradicts that B is nonzero.</p><p><strong>Step 2:</strong> Therefore A cannot be invertible, so det(A) = 0. Similarly, B cannot be invertible, so det(B) = 0.</p><p><strong>Step 3:</strong> Since both A and B are nonzero but singular matrices, both must have rank less than n. The column space of A lies entirely in the null space of B, which is consistent with AB = O.</p><p><strong>Step 4:</strong> We conclude: Both A and B are singular matrices (det(A) = 0 and det(B) = 0), and rank(A) + rank(B) ≤ n.</p><p>∴ Answer: A</p>
Correct Answer: A