<p>Two n × n square matrices A and B are said to be similar if there exists a non-singular matrix P such that P<sup>−1</sup> A P = B</p><p>Match Column I with Column II:</p><p><b>Column I:</b><br>(A) If A and B are similar matrices such that det(AB) = 0, then<br>(B) If A and B are two non-singular matrices, then<br>(C) If A and B are similar matrices such that det(A) = 1, then<br>(D) If A and B are two non-triangular matrices, then</p><p><b>Column II:</b><br>(p) det(A) = det(B) ≠ 0<br>(q) det(AB) = det(BA)<br>(r) det(A) + det(B) = 2<br>(s) det(A) = det(B) = 0</p>
Step-by-Step Solution
Key Concept: Similar matrices have identical determinants (det(P⁻¹AP) = det(A)), and determinant is multiplicative and commutative for the product of any two matrices (det(AB) = det(BA) = det(A)·det(B)). Use these properties along with given constraints to match statements.
<p><strong>Step 1: Analyze Statement A</strong></p><p>Given: A and B are similar with det(AB) = 0</p><p>Since A ~ B, we have P⁻¹AP = B, so det(B) = det(A)</p><p>det(AB) = det(A)·det(B) = [det(A)]² = 0</p><p>Therefore det(A) = det(B) = 0 → <strong>A → s</strong></p><p><strong>Step 2: Analyze Statement B</strong></p><p>Given: A and B are non-singular matrices</p><p>For ANY square matrices: det(AB) = det(A)·det(B) = det(B)·det(A) = det(BA)</p><p>This property holds regardless of whether matrices are similar → <strong>B → q</strong></p><p><strong>Step 3: Analyze Statement C</strong></p><p>Given: A and B are similar with det(A) = 1</p><p>Since A ~ B: det(B) = det(A) = 1</p><p>Therefore det(A) + det(B) = 1 + 1 = 2 → <strong>C → r</strong></p><p><strong>Step 4: Analyze Statement D</strong></p><p>Given: A and B are two non-triangular matrices</p><p>Being non-triangular is not a restrictive property for determinants. For ANY square matrices A and B:</p><p>det(AB) = det(A)·det(B) = det(B)·det(A) = det(BA)</p><p>This holds regardless of triangular structure → <strong>D → q</strong></p><p><strong>∴ Answer: A→s, B→q, C→r, D→q</strong></p>
Correct Answer: A→s, B→q, C→r, D→q