(A∗)∗=∣A∣n−2⋅A(A^*)^*=|A|^{n-2}·A(A∗)∗=∣A∣n−2⋅A(其中A可逆，A∗A^*A∗代表矩阵A的伴随矩阵，∣A∣|A|∣A∣表示矩阵A的行列式det(A)det(A)det(A))
证明如下：
∣A∗∣=∣∣A∣A−1∣=∣A∣n⋅∣A∣−1=∣A∣n−1|A^*|=\big||A|A^{-1}\big|=|A|^n·|A|^{-1}=|A|^{n-1}∣A∗∣=∣∣∣A∣A−1∣∣=∣A∣n⋅∣A∣−1=∣A∣n−1
(A∗)∗=∣A∗∣⋅(A∗)−1=∣A∣n−1∗(A∗)−1(A^*)^*=|A^*|·(A^*)^{-1}=|A|^{n-1}*(A^*)^{-1}(A∗)∗=∣A∗∣⋅(A∗)−1=∣A∣n−1∗(A∗)−1
A∗=∣A∣⋅A−1⇒(A∗)−1=A∣A∣A^*=|A|·A^{-1}\Rightarrow (A^*)^{-1}=\displaystyle\frac{A}{|A|}A∗=∣A∣⋅A−1⇒(A∗)−1=∣A∣A
由2，3得(A∗)∗=∣A∣n−1⋅A∣A∣=∣A∣n−2∗A(A^*)^*=|A|^{n-1}·\displaystyle\frac{A}{|A|}=|A|^{n-2}*A(A∗)∗=∣A∣n−1⋅∣A∣A=∣A∣n−2∗A
