Question:

<p><strong>Step 1: Analyze matrix $P$</strong></p><p>$P$ is upper triangular with diagonal entries all equal to $1$. We can write $P = I + N$ where $N$ is nilpotent (strictly upper triangular).</p><p><strong>Step 2: Find pattern for $P^n$</strong></p><p>Since $P = I + N$ with $N = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, we compute:</p><p>$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ (nilpotent index $2$)</p><p>By binomial expansion: $P^n = I + nN$</p><p>Therefore $A = P^{2018} = I + 2018N = \begin{bmatrix} 1 & 2018 & 0 \\ 0 & 1 & 0 \\ 0 & 2018 & 1 \end{bmatrix}$</p><p><strong>Step 3: Analyze matrix $B = QPQ^T$</strong></p><p>Since $Q$ is orthogonal, $QPQ^T$ is similar to $P$ via an orthogonal transformation. This preserves:</p><ul><li>Eigenvalues (all equal $1$)</li><li>Trace ($\text{tr}(B) = \text{tr}(P) = 3$)</li><li>Determinant ($\det(B) = \det(P) = 1$)</li><li>Rank ($\text{rank}(B) = \text{rank}(P) = 3$)</li></ul><p><strong>Step 4: Verify each option</strong></p><p><strong>(a) Trace of $A$ is $3$:</strong> Diagonal entries of $A$ are $1, 1, 1$, so $\text{tr}(A) = 3$ ✓</p><p><strong>(b) $Q^T B^{2018} Q = A$:</strong> Since $B = QPQ^T$, we have $B^{2018} = QPQ^T \cdot QPQ^T \cdots QPQ^T = QP^{2018}Q^T = QAQ^T$. Therefore $Q^T B^{2018} Q = Q^T QAQ^T Q = A$ ✓</p><p><strong>(c) $\det(B^5) = 1$:</strong> Since $\det(B) = \det(P) = 1$ (product of diagonal entries of upper triangular matrix), we have $\det(B^5) = (\det(B))^5 = 1^5 = 1$ ✓</p><p><strong>(d) $\det(\text{adj}(A)) = \det(\text{adj}(B))$:</strong> For a $3 \times 3$ matrix $M$, $\det(\text{adj}(M)) = (\det(M))^{3-1} = (\det(M))^2$. Since $\det(A) = 1$ and $\det(B) = 1$, we have $\det(\text{adj}(A)) = 1$ and $\det(\text{adj}(B)) = 1$, so they are equal ✓</p><p><strong>Answer: (a), (b), (c), (d)</strong></p>