<p>Let \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\). Then</p>
<p>\(A^2 - 4A - 5I_3 = O\)</p>
<p>\(A^{-1} = \frac{1}{5}(A - 4I_3)\)</p>
<p>\(A^3\) is not invertible</p>
<p>\(A^2\) is invertible</p>
Step-by-Step Solution
Key Concept: Find eigenvalues by solving det(A - λI) = 0, then use eigenvalue properties to determine which statements about A are true (rank, diagonalizability, trace relationships).
<p><strong>Step 1:</strong> Find the characteristic polynomial.</p><p>det(A - λI) = det⎛⎜⎝1-λ 2 2 ⎞⎟⎠ = (1-λ)[(1-λ)² - 4] - 2[2(1-λ) - 4] + 2[4 - 2(1-λ)]</p><p> 2 1-λ 2</p><p> 2 2 1-λ</p><p>Simplifying: (1-λ)[(1-λ)² - 4] - 2[2 - 2λ - 4] + 2[4 - 2 + 2λ]</p><p>= (1-λ)(λ² - 2λ - 3) - 2(-2 - 2λ) + 2(2 + 2λ)</p><p>= (1-λ)(λ-3)(λ+1) + 4(1 + λ) + 4(1 + λ)</p><p>= -(λ-5)(λ+1)²</p><p><strong>Step 2:</strong> The eigenvalues are λ₁ = 5 (multiplicity 1) and λ₂ = -1 (multiplicity 2).</p><p><strong>Step 3:</strong> Verify statements:</p><p>• <strong>A:</strong> Rank of A = 3 if all eigenvalues ≠ 0. Since 5 and -1 are both nonzero, rank(A) = 3. ✓</p><p>• <strong>B:</strong> Trace(A) = 1 + 1 + 1 = 3 = sum of eigenvalues = 5 + (-1) + (-1) = 3. ✓</p><p>• <strong>C:</strong> det(A) = product of eigenvalues = 5·(-1)·(-1) = 5 ≠ -1. ✗</p><p>• <strong>D:</strong> A is diagonalizable if geometric multiplicity equals algebraic multiplicity for λ = -1. Since the eigenspace dimension for λ = -1 is 2, A is diagonalizable. ✓</p><p>∴ Answer: A, B, D</p>
Correct Answer: A,B,D