Determinant of anti-diagonal permutation matrix

Consider a $5\times5$ matrix $P=(5,4,3,2,1)$ which means it has anti-diagonal entries of $1$'s.

If we calculate $\det P$ using the theorem "The determinants changes sign when two rows are exchange", then it is : 1 (making it identity) But if we use cofactors then the answer is : $-1$
Is that possible? Which one is true?

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2 Answers

Now that is not possible. You made an error using cofactors, we have \begin{align*} \det P &= \det\begin{pmatrix} 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\end{pmatrix}\\ &= (-1)^{1+5} \det \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\end{pmatrix}\\ &= 1 \cdot (-1)^{1+4} \det \begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\end{pmatrix}\\ &= -1 \cdot (-1)^{1+3} \det \begin{pmatrix} 0 & 1\\ 1 &0\end{pmatrix}\\ &= -1 \cdot (-1)\\ &= 1. \end{align*}

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The (1,5)-th minor $M_{15}$ of $P$ (i.e. the determinant of the 4x4 submatrix obtained by removing the first row and the fifth column of $P$) is $1$, so the (1,5)-th cofactor is $C_{15}=(-1)^{1+5}M_{15}=1$ and in turn, if you evaluate $\det P$ using cofactors, you should get the answer as $1$ too, not $-1$.

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