Question

If \( \operatorname{adj} B = A, \ |P|=|Q|=1 \), then \[ \operatorname{adj}(Q^{-1} B P^{-1}) = \ ? \]

• A. \( PQ \)
• B. \( QAP \)
• C. \( PAQ \)
• D. \( PA^{-1}Q \)

Hint:

Remember: \begin{itemize} \item Adjoint reverses multiplication order. \item Determinant 1 matrices simplify adjoint to inverse. \end{itemize}

Answer 1

The Correct Option is C

Concept: Key properties: \begin{itemize} \item \( \operatorname{adj}(ABC) = \operatorname{adj}(C)\operatorname{adj}(B)\operatorname{adj}(A) \) \item \( \operatorname{adj}(M^{-1}) = (\operatorname{adj} M)^{-1} \) \item If \( |M|=1 \Rightarrow \operatorname{adj} M = M^{-1} \) \end{itemize} Step 1: {\color{red}Apply adjoint of product.} \[ \operatorname{adj}(Q^{-1}BP^{-1}) = \operatorname{adj}(P^{-1})\operatorname{adj}(B)\operatorname{adj}(Q^{-1}) \] Step 2: {\color{red}Use given information.} Given: \[ \operatorname{adj}B = A \] Also \( |P|=|Q|=1 \Rightarrow \operatorname{adj}P = P^{-1}, \ \operatorname{adj}Q = Q^{-1} \). Hence: \[ \operatorname{adj}(P^{-1}) = P, \quad \operatorname{adj}(Q^{-1}) = Q \] Step 3: {\color{red}Substitute.} \[ \operatorname{adj}(Q^{-1}BP^{-1}) = P \cdot A \cdot Q \]