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

Step 1: Simplify \(A^2B+A^4B\).& nbsp;

Since

\[ A= \begin{bmatrix} 0 & amp; 0 & amp; 1\\ 0 & amp; 1 & amp; 0\\ 1 & amp; 0 & amp; 0 \end{bmatrix}, \]

we have

\[ A^2=I. \]

Hence,

\[ A^4=(A^2)^2=I. \]

Therefore,

\[ A^2B+A^4B = B+B = 2B. \]

Thus,

\[ (2B)^{-1} = \frac{1}{2} \begin{bmatrix} 1 & amp; 2 & amp; -1\\ 0 & amp; 1 & amp; 2\\ 1 & amp; 0 & amp; 1 \end{bmatrix}. \]

Multiplying both sides by \(2\),

\[ B^{-1} = \begin{bmatrix} 1 & amp; 2 & amp; -1\\ 0 & amp; 1 & amp; 2\\ 1 & amp; 0 & amp; 1 \end{bmatrix}. \]

Step 2: Find \(B\).

Taking the inverse,

\[ B = \left( \begin{bmatrix} 1 & amp; 2 & amp; -1\\ 0 & amp; 1 & amp; 2\\ 1 & amp; 0 & amp; 1 \end{bmatrix} \right)^{-1}. \]

The determinant of the matrix is

\[ \begin{vmatrix} 1 & amp; 2 & amp; -1\\ 0 & amp; 1 & amp; 2\\ 1 & amp; 0 & amp; 1 \end{vmatrix} = 5\neq0. \]

Hence,

\[ B = \frac{1}{5} \begin{bmatrix} 1 & amp; -2 & amp; 5\\ 2 & amp; 2 & amp; -2\\ -1 & amp; 2 & amp; 1 \end{bmatrix}. \]

Therefore,

\[ x=\frac{1}{5},\; y=-\frac{2}{5},\; z=1, \] \[ l=\frac{2}{5},\; m=\frac{2}{5},\; n=-\frac{2}{5}, \] \[ p=-\frac{1}{5},\; q=\frac{2}{5},\; r=\frac{1}{5}. \]

Step 3: Evaluate the required expression.

\[ xyz = \frac{1}{5} \left(-\frac{2}{5}\right) (1) = -\frac{2}{25}. \] \[ lmn = \frac{2}{5}\cdot \frac{2}{5}\cdot \left(-\frac{2}{5}\right) = -\frac{8}{125}. \] \[ pqr = -\frac{1}{5}\cdot \frac{2}{5}\cdot \frac{1}{5} = -\frac{2}{125}. \]

Hence,

\[ xyz-lmn-pqr = -\frac{2}{25} +\frac{8}{125} +\frac{2}{125} = -\frac{10}{125} +\frac{10}{125} = 0. \]

Therefore,

\[ \boxed{0}. \]