Question

If \[ A=\begin{bmatrix} 1 & 2 & i\\ 1 & 1 & 1\\ 1 & 1 & 0 \end{bmatrix}, \] then \( [\operatorname{adj}(\operatorname{adj}A)]^{-1} \) is

• A. \( A^2 \)
• B. \( 2A \)
• C. \( A^{-1} \)
• D. \( I \)

Hint:

Remember the identity: \( \operatorname{adj}(\operatorname{adj}A) = (\det A)^{n-2}A \) for an \( n\times n \) matrix.

Answer 1

The Correct Option is C

Step 1: Recall the adjugate property.
For an \( n\times n \) matrix, \[ \operatorname{adj}(\operatorname{adj}A) = (\det A)^{\,n-2}A \] Here \( n=3 \).

Step 2: Apply the formula.
\[ \operatorname{adj}(\operatorname{adj}A) = (\det A)A \]

Step 3: Take inverse.
\[ [\operatorname{adj}(\operatorname{adj}A)]^{-1} = \frac{1}{\det A}A^{-1} \] Since \( \det A \neq 0 \), \[ [\operatorname{adj}(\operatorname{adj}A)]^{-1}=A^{-1} \]

Step 4: Conclusion.
\[ \boxed{A^{-1}} \]