<p>Let \(A = \begin{bmatrix} a & b & 1 \\ 2 & 1 & 3 \\ 1 & c & 2 \end{bmatrix}\) and \(A^{-1} = (5A - A^2)\), then:</p>
Step-by-Step Solution
Key Concept: Use the condition A⁻¹ = 5A - A² to derive A² - 5A + I = 0, then apply the Cayley-Hamilton theorem to find relationships between a, b, c through the characteristic equation and trace/determinant constraints.
<p><strong>Step 1:</strong> From A⁻¹ = 5A - A², multiply both sides by A: I = 5A² - A³, giving A³ - 5A² + I = 0 (or equivalently A² - 5A + I = 0 after multiplying A⁻¹ by A from right).</p><p><strong>Step 2:</strong> The condition A⁻¹ = 5A - A² implies AA⁻¹ = I, so A(5A - A²) = I, which gives 5A² - A³ = I, or <strong>A² - 5A + I = 0</strong>.</p><p><strong>Step 3:</strong> This is the characteristic equation, so trace(A) = 5 and det(A) = 1. From trace: a + 1 + 2 = 5, therefore <strong>a = 2</strong>.</p><p><strong>Step 4:</strong> Calculate det(A) = a(2-3c) - b(4-3) + 1(2c-1) = 2(2-3c) - b + 2c - 1 = 4 - 6c - b + 2c - 1 = 3 - 4c - b = 1. Thus <strong>b + 4c = 2</strong>.</p><p><strong>Step 5:</strong> Verify A² - 5A + I = 0 using a=2 and b+4c=2. Check that the resulting matrix equation holds for any b,c satisfying b + 4c = 2, with a = 2. Options B, C, D correspond to valid (a,b,c) triples satisfying these constraints.</p><p>∴ Answer: B,C,D</p>
Correct Answer: B,C,D