Matrices & Determinants
Cayley-Hamilton theorem
Grade None
Question:
<p>Let \( A \) be a \( 3 \times 3 \) matrix such that \( A^2 - 5A + 7I = 0 \).</p><p><strong>Statement-I:</strong> \( A^{-1} = \dfrac{1}{7}(5I - A) \).</p><p><strong>Statement-II:</strong> The polynomial \( A^3 - 2A^2 - 3A + I \) can be reduced to \( 5(A - 4I) \).</p><p>Then:</p>
<p>Both statements are true.</p>
<p>Both statements are false.</p>
<p>Statement-I is true, but Statement-II is false.</p>
<p>Statement-I is false, but Statement-II is true.</p>
Step-by-Step Solution
Key Concept: Use the given matrix equation A² - 5A + 7I = 0 to find A⁻¹ by rearranging as A(A - 5I) = -7I, and express higher powers of A in terms of lower powers using the minimal polynomial relation.
<p><strong>Step 1: Verify Statement-I (Finding A⁻¹)</strong></p><p>From A² - 5A + 7I = 0, rearrange as:</p><p>A² - 5A = -7I</p><p>A(A - 5I) = -7I</p><p>A · 1/(-7)(A - 5I) = I</p><p>Therefore: A⁻¹ = 1/7(5I - A) ✓</p><p><strong>Statement-I is TRUE</strong></p><p><strong>Step 2: Verify Statement-II (Reduce A³ - 2A² - 3A + I)</strong></p><p>From A² - 5A + 7I = 0, we have: A² = 5A - 7I</p><p>Find A³:</p><p>A³ = A · A² = A(5A - 7I) = 5A² - 7A</p><p>Substitute A² = 5A - 7I:</p><p>A³ = 5(5A - 7I) - 7A = 25A - 35I - 7A = 18A - 35I</p><p><strong>Step 3: Substitute into the expression</strong></p><p>A³ - 2A² - 3A + I</p><p>= (18A - 35I) - 2(5A - 7I) - 3A + I</p><p>= 18A - 35I - 10A + 14I - 3A + I</p><p>= (18 - 10 - 3)A + (-35 + 14 + 1)I</p><p>= 5A - 20I</p><p>= 5(A - 4I) ✓</p><p><strong>Statement-II is TRUE</strong></p><p><strong>∴ Answer: C (Both Statement-I and Statement-II are TRUE)</strong></p>
Correct Answer: C