伴随矩阵最基本的恒等式为: $$ A(A^*) = |A|E $$

怎么证明呢? 参考了这里

直接从定义出发来证明。 以一个 3x3 的具体行列式为例。

$$ A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{vmatrix} $$

根据定义，A 的伴随矩阵 A^*为每个元素的代数余子数按列排组成的矩阵。

$$ A^* = \begin{vmatrix} A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33} \\
\end{vmatrix} $$

所以:

$$ A * A^* = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{vmatrix} \begin{vmatrix} A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33} \\
\end{vmatrix} = \\
\begin{vmatrix} a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} & a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} & a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} \\
a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13} & a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} & a_{21}A_{31} + a_{22}A_{32} + a_{23}A_{33} \\
a_{31}A_{11} + a_{32}A_{12} + a_{33}A_{13} & a_{31}A_{21} + a_{32}A_{22} + a_{33}A_{23} & a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33} \\
\end{vmatrix} = \\
\begin{vmatrix} |A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A| \\
\end{vmatrix} = |A|E $$