Question

If \[ A=\operatorname{Adj}\!\left(-2\left(\operatorname{Adj}(P^{-1})\right)\right) \] and \[ P= \begin{bmatrix} 1& 2& 1 1& 0& 1 0& 1& 2 \end{bmatrix}, \] then \(|A|=\)

• A. \(-\dfrac14\)
• B. \(\dfrac14\)
• C. \(\dfrac34\)
• D. \(\dfrac23\)

Hint:

For an \(n\times n\) matrix, \[ \boxed{ |\operatorname{Adj}(A)|=|A|^{\,n-1} } \] and \[ \boxed{ |kA|=k^n|A|. } \] For a \(3\times3\) matrix, \[ |\operatorname{Adj}(A)|=|A|^2. \]

Answer 1

The Correct Option is B

Step 1: Find the determinant of \(P\).& nbsp;

\[ |P| = \begin{vmatrix} 1 & amp; 2 & amp; 1\\ 1 & amp; 0 & amp; 1\\ 0 & amp; 1 & amp; 2 \end{vmatrix}. \]

Expanding along the first row,

\[ |P| = 1 \begin{vmatrix} 0 & amp; 1\\ 1 & amp; 2 \end{vmatrix} - 2 \begin{vmatrix} 1 & amp; 1\\ 0 & amp; 2 \end{vmatrix} + 1 \begin{vmatrix} 1 & amp; 0\\ 0 & amp; 1 \end{vmatrix}. \] \[ |P| = (-1)-4+1 = -4. \]

Hence,

\[ \boxed{|P|=-4.} \]

Step 2: Find the determinant of \(\operatorname{Adj}(P^{-1})\).

Since \(P\) is a \(3\times3\) matrix,

\[ |\operatorname{Adj}(M)|=|M|^{2}. \]

Also,

\[ |P^{-1}|=\frac{1}{|P|}=-\frac{1}{4}. \]

Therefore,

\[ \left|\operatorname{Adj}(P^{-1})\right| = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}. \]

Step 3: Find \(|A|\).

Let

\[ B=-2\left(\operatorname{Adj}(P^{-1})\right). \]

Since \(B\) is a \(3\times3\) matrix,

\[ |B| = (-2)^3 \cdot \frac{1}{16} = -\frac{1}{2}. \]

Again,

\[ A=\operatorname{Adj}(B). \]

Hence,

\[ |A| = |B|^{2} = \left(-\frac{1}{2}\right)^2 = \boxed{\frac{1}{4}}. \]

Therefore, the correct option is

\[ \boxed{(B)}. \]