If \(K\) is a commutative ring with identity and \(A\) is the matrix over \(K\) given by \[ A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \] show that \(\det A = 0\).
Solution
Because the determinant is unchanged by transpose, \[ \det A=\det(A^t). \] Here \(A^t=-A\). By multilinearity in the three rows, \[ \det(-A)=(-1)^3\det A=-\det A. \] Thus \(\det A=-\det A\), so \(2\det A=0\). This argument alone is insufficient in characteristic \(2\), so compute without division: expansion along the first row gives \[ \det A=-a\begin{vmatrix}-a&c\\-b&0\end{vmatrix} +b\begin{vmatrix}-a&0\\-b&-c\end{vmatrix} =-abc+abc=0. \] Hence \(\det A=0\) over every commutative ring \(K\).