Matrices & Determinants
Adjoint and Inverse of a Matrix
Grade 12
Question:
<p>Let \(A\) be a square matrix of order 3 satisfies the relation \(A^3 - 6A^2 + 7A - 8I = O\) and \(B = A - 2I\). Also, det.\(A = 8\), then</p>
<p>\(\det.(\text{adj.}(I - 2A^{-1})) = \dfrac{25}{16}\)</p>
<p>\(\text{adj.}\left(\left(\dfrac{B}{2}\right)^{-1}\right) = \dfrac{B}{10}\)</p>
<p>\(\det.(\text{adj.}(I - 2A^{-1})) = \dfrac{75}{32}\)</p>
<p>\(\text{adj.}\left(\left(\dfrac{B}{2}\right)^{-1}\right) = \dfrac{2B}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the given matrix equation A³ - 6A² + 7A - 8I = O to find a relation involving B = A - 2I, then substitute A = B + 2I to derive a polynomial equation in B that reveals its determinant or key property.
<p><strong>Step 1:</strong> Given: A³ - 6A² + 7A - 8I = O and B = A - 2I, so A = B + 2I</p><p><strong>Step 2:</strong> Substitute A = B + 2I into the relation:</p><p>(B + 2I)³ - 6(B + 2I)² + 7(B + 2I) - 8I = O</p><p><strong>Step 3:</strong> Expand each term:</p><p>(B + 2I)³ = B³ + 6B² + 12B + 8I</p><p>-6(B + 2I)² = -6B² - 24B - 24I</p><p>7(B + 2I) = 7B + 14I</p><p><strong>Step 4:</strong> Combine all terms:</p><p>B³ + 6B² + 12B + 8I - 6B² - 24B - 24I + 7B + 14I - 8I = O</p><p>B³ + (6 - 6)B² + (12 - 24 + 7)B + (8 - 24 + 14 - 8)I = O</p><p>B³ - 5B + (-10)I = O</p><p><strong>Step 5:</strong> This means B³ - 5B + 10I = O, or equivalently B³ = 5B - 10I</p><p>Taking determinants: det(B³) = det(5B - 10I)</p><p>[det(B)]³ = det(5B - 10I), which can be used to find det(B) or verify constraints from det(A) = det(B + 2I) = 8</p><p>∴ Answer: A</p>
Correct Answer: A