Question

Let \(A\) be a \(3\times3\) matrix such that \[ \det(A)=-1. \] If \[ B^{-1}=Adj\!\left(A\,Adj(A^2)\right), \] then find \[ \det((\det A)B). \]

• A. \(|A|\)
• B. \(B\)
• C. \(|A+B|\)
• D. \(|A-B|\)

Hint:

For \(3\times3\) matrices, \[ Adj(A)=|A|A^{-1} \] is often the fastest route to simplify complicated determinant expressions.

Answer 1

The Correct Option is A

Concept: For an \(n\times n\) matrix, \[ Adj(A)=|A|A^{-1}. \] Also, \[ |Adj(A)|=|A|^{\,n-1}. \]

Step 1: Evaluate \(Adj(A^2)\).
Since \[ |A|=-1, \] \[ |A^2|=1. \] Therefore \[ Adj(A^2)=A^{-2}. \]

Step 2: Simplify the expression.
\[ A\,Adj(A^2) = A\cdot A^{-2} = A^{-1}. \] Hence \[ B^{-1}=Adj(A^{-1}). \]

Step 3: Find \(Adj(A^{-1})\).
For a \(3\times3\) matrix, \[ Adj(A^{-1}) = |A^{-1}|(A^{-1})^{-1}. \] Since \[ |A^{-1}|=-1, \] \[ Adj(A^{-1}) = -A. \] Thus \[ B^{-1}=-A. \] Hence \[ B=-A^{-1}. \]

Step 4: Compute the determinant.
\[ \det(B) = \det(-A^{-1}) = (-1)^3\det(A^{-1}) = -(-1) = 1. \] Now \[ \det((\det A)B) = (-1)^3\det(B) = -1. \] Since \[ |A|=-1, \] \[ \boxed{\det((\det A)B)=|A|} \] Hence, \[ \boxed{\text{Correct Option (1)}} \]