Matrices & Determinants
Orthogonal matrices, adjoint, and determinant properties
GRB_1000_MCQ
Grade Class 12
Question:
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 <b>correct</b>?
Trace of matrix $A$ is 3
$Q^T B^{2018} Q = A$
$\det(B^5) = 1$
$\det(adj(A)) = \det(adj(B))$
Step-by-Step Solution
Step 1: Find $\det(P)$. $P$ is upper/lower triangular-like; $\det(P) = 1\cdot1\cdot1 = 1$.
Step 2: Trace of $A = P^{2018}$. Since $P$ is a unipotent matrix (eigenvalues all 1), $A = P^{2018}$ also has all eigenvalues equal to 1, so trace$(A) = 3$. ✓ Option (1) is correct.
Step 3: Verify $Q^T B^{2018} Q = A$. Since $B = QPQ^T$, we have $B^{2018} = QP^{2018}Q^T = QAQ^T$. Then:
$$Q^T B^{2018} Q = Q^T(QAQ^T)Q = A$$
Option (2) is correct. ✓
Step 4: Compute $\det(B^5)$. Since $\det(B) = \det(Q)\det(P)\det(Q^T) = (\pm1)(1)(\pm1) = 1$, so $\det(B^5) = (\det B)^5 = 1$. ✓ Option (3) is correct.
Step 5: Verify $\det(adj(A)) = \det(adj(B))$. For a $3\times3$ matrix $M$, $\det(adj(M)) = (\det M)^{3-1} = (\det M)^2$. Since $\det(A) = \det(P^{2018}) = 1$ and $\det(B) = 1$:
$$\det(adj(A)) = 1 = \det(adj(B))$$
Option (4) is correct. ✓
Correct Answer: 1, 2, 3, 4