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: Link cofactors to the adjugate.
The matrix of cofactors $Q$ has the property that its transpose is the adjugate of $P$, so $Q^T=\operatorname{adj}(P)$.

Step 2: Determinant is transpose safe.
Since $\det(Q)=\det(Q^T)$, we have $\det(Q)=\det(\operatorname{adj}(P))$.

Step 3: Recall the adjugate rule.
For an $n\times n$ matrix, $\det(\operatorname{adj}(P))=(\det P)^{\,n-1}$.

Step 4: Put $n=5$.
\[ \det(Q)=(\det P)^{4} \]

Step 5: Use $\det P=2$.
\[ \det(Q)=2^4=16 \]
\[ \boxed{16.0} \]