<p><strong>996.</strong> Let \(P\) be a \(2 \times 2\) matrix such that \(P\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}-1\\2\end{bmatrix}\) and \(P^2\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}\). If \(x_1\) and \(x_2\) are two values of \(x\) for which \(|P - xI| = 0\), where \(I\) is an identity matrix of order 2, then find the value of \(x_1^2 + x_2^2\).</p>
Step-by-Step Solution
Key Concept: Use the given conditions to find the characteristic polynomial of P by recognizing that if v is an eigenvector with eigenvalue λ, then Pv = λv. The eigenvalues satisfy |P - xI| = 0, and we can use Cayley-Hamilton theorem combined with the given conditions to find x₁² + x₂².
<p><strong>Step 1: Identify the relationship with eigenvalues.</strong></p><p>Let v = [1, -1]ᵀ. We're given:</p><p>Pv = [-1, 2]ᵀ and P²v = [1, 0]ᵀ</p><p>Note: P²v = P(Pv) = P[-1, 2]ᵀ = [1, 0]ᵀ</p><p><strong>Step 2: Use the characteristic polynomial approach.</strong></p><p>Since x₁ and x₂ are eigenvalues of P, the characteristic polynomial is:</p><p>|P - xI| = (x - x₁)(x - x₂) = x² - (x₁ + x₂)x + x₁x₂</p><p>By Cayley-Hamilton theorem: P² - (x₁ + x₂)P + x₁x₂·I = 0</p><p><strong>Step 3: Apply the condition P²v = [1, 0]ᵀ.</strong></p><p>From Cayley-Hamilton: P²v = (x₁ + x₂)Pv - x₁x₂·v</p><p>Substituting: [1, 0]ᵀ = (x₁ + x₂)[-1, 2]ᵀ - x₁x₂[1, -1]ᵀ</p><p>Component-wise:</p><p>1 = -(x₁ + x₂) - x₁x₂ ... (i)</p><p>0 = 2(x₁ + x₂) + x₁x₂ ... (ii)</p><p><strong>Step 4: Solve for trace and determinant.</strong></p><p>From equation (ii): x₁x₂ = -2(x₁ + x₂)</p><p>Substituting into equation (i):</p><p>1 = -(x₁ + x₂) - [-2(x₁ + x₂)] = -(x₁ + x₂) + 2(x₁ + x₂) = (x₁ + x₂)</p><p>Therefore: x₁ + x₂ = 1</p><p>And: x₁x₂ = -2(1) = -2</p><p><strong>Step 5: Calculate x₁² + x₂².</strong></p><p>Using the identity: x₁² + x₂² = (x₁ + x₂)² - 2x₁x₂</p><p>x₁² + x₂² = (1)² - 2(-2) = 1 + 4 = 5</p><p><strong>∴ Answer: 5</strong></p>
Correct Answer: 5