<p>Given \((A - 3I)(A - 5I) = 0\) for a matrix \(A\), which gives \(\alpha A + \beta A^{-1} = 4I\). Then \(\alpha + \beta\) equals:</p>
Step-by-Step Solution
Key Concept: From (A - 3I)(A - 5I) = 0, we get A² - 8A + 15I = 0, which means A² = 8A - 15I. Dividing by A gives A = 8I - 15A⁻¹, leading to αA + βA⁻¹ = 4I by rearrangement.
<p><strong>Step 1:</strong> Expand (A - 3I)(A - 5I) = 0</p><p>A² - 5A - 3A + 15I = 0</p><p>A² - 8A + 15I = 0</p><p><strong>Step 2:</strong> Since A is invertible (A⁻¹ exists), divide the equation by A</p><p>A - 8I + 15A⁻¹ = 0</p><p><strong>Step 3:</strong> Rearrange to match the form αA + βA⁻¹ = 4I</p><p>A + 15A⁻¹ = 8I</p><p>Multiply by (1/2): (1/2)A + (15/2)A⁻¹ = 4I</p><p><strong>Step 4:</strong> Identify coefficients</p><p>α = 1, β = 15 (from A + 15A⁻¹ = 8I)</p><p>Or scale-invariantly: α = 1/2, β = 15/2</p><p>∴ α + β = 1 + 15 = <strong>16</strong> or with scaling α + β = 1/2 + 15/2 = <strong>8</strong></p><p><strong>Note:</strong> The answer 8 comes from recognizing that A + 15A⁻¹ = 8I directly gives us the sum of coefficients when normalized: coefficient of A is 1 and coefficient of A⁻¹ is 15, but the RHS constant is 8, which is the answer sought.</p>
Correct Answer: 8