Question

If \[ A= \begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix}, \qquad B= \begin{bmatrix} x & y & z\\ l & m & n\\ p & q & r \end{bmatrix}, \] and \[ \left(A^2B+A^4B\right)^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 2 & -1\\ 0 & 1 & 2\\ 1 & 0 & 1 \end{bmatrix}, \] then the value of \[ xyz-lmn-pqr \] is

• A. \(\dfrac{37}{6^3}\)
• B. \(\dfrac{45}{2^3}\)
• C. \(1\)
• D. \(0\)

Hint:

If a permutation matrix satisfies \[ A^2=I, \] then automatically \[ \boxed{A^{2k}=I} \] for every positive integer \(k\). This greatly simplifies matrix expressions involving even powers.

Answer 1

The Correct Option is D