Question

Let \(P\) be a \(5\times 5\) real matrix with \(\det(P)=2\). Let \(Q\) be the matrix of cofactors of \(P\). Then \(\det(Q)=\underline{}\) rounded off to one decimal place.

Hint:

For an \(n\times n\) matrix \(A\), \(\det(\operatorname{adj}(A))=(\det A)^{n-1}\).

Answer 1

Correct Answer: 16

Step 1: Recall cofactor matrix relation.
The transpose of the cofactor matrix is the adjugate matrix.
\[ \operatorname{adj}(P)=Q^T \]

Step 2: Use determinant property.
Since determinant of a matrix and its transpose are equal,
\[ \det(Q)=\det(Q^T) \]

Step 3: Replace \(Q^T\) by adjugate matrix.
\[ \det(Q)=\det(\operatorname{adj}(P)) \]

Step 4: Use formula for adjugate determinant.
For an \(n\times n\) matrix \(P\),
\[ \det(\operatorname{adj}(P))=(\det P)^{n-1} \]

Step 5: Substitute \(n=5\).
\[ \det(Q)=(\det P)^4 \]

Step 6: Substitute \(\det(P)=2\).
\[ \det(Q)=2^4 \]

Step 7: Final answer.
\[ \det(Q)=16 \] \[ \boxed{16.0} \]