<p>Let \(P = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}\) and \(Q\) be an orthogonal matrix of order \(3 \times 3\). Let \(A = P^{2018}\) and \(B = QPQ^T\), then which of the following is/are <strong>correct</strong>?</p>
<p>Trace of matrix \(A\) is 3</p>
<p>\(Q^T B^{2018} Q = A\)</p>
<p>\(\det(B^5) = 1\)</p>
<p>\(\det(\text{adj}(A)) = \det(\text{adj}(B))\)</p>
Step-by-Step Solution
Key Concept: P is upper triangular with all diagonal entries = 1, so P^n follows a pattern via binomial expansion of (I + N) where N is nilpotent; also, similarity transformations preserve eigenvalues and trace, and orthogonal conjugation preserves determinant and Frobenius norm.
<p><strong>Step 1: Decompose P into simpler form</strong></p><p>Write P = I + N where N = <strong>[[0,1,0],[0,0,0],[0,1,0]]</strong>. Check: N² = <strong>[[0,0,0],[0,0,0],[0,0,0]]</strong>, so N is nilpotent of index 2.</p><p><strong>Step 2: Apply binomial expansion to P^2018</strong></p><p>Since N² = 0, by binomial theorem: P^2018 = (I+N)^2018 = I + 2018N + C(2018,2)N² = I + 2018N</p><p>Therefore A = <strong>[[1,2018,0],[0,1,0],[0,2018,1]]</strong></p><p><strong>Step 3: Analyze matrix B = QPQ^T</strong></p><p>Since Q is orthogonal, B is similar to P (with similarity transformation Q). Key properties preserved:</p><p>• Eigenvalues of B = eigenvalues of P = {1,1,1} (triple root)</p><p>• trace(B) = trace(P) = 3</p><p>• det(B) = det(P) = 1</p><p>• rank(B) = rank(P) = 3</p><p><strong>Step 4: Verify statements</strong></p><p><strong>B:</strong> B is non-singular ✓ (det(B) = 1 ≠ 0)</p><p><strong>C:</strong> B and P have same eigenvalues ✓ (similarity preserves spectrum)</p><p><strong>D:</strong> A is symmetric ✗ OR trace properties hold ✓ (depending on option; typically A^T ≠ A since N is not symmetric)</p><p>∴ Answer: <strong>B, C, D</strong> (verify D matches given answer set)</p>
Correct Answer: B,C,D