上述性质可以推导伴随矩阵的性质 A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E
设 A = ( a i j ) A=(a_{ij}) A=(aij), B = A A ∗ = ( b i j ) B=AA^*=(b_{ij}) B=AA∗=(bij)
A ∗ = ( c 11 c 12 ⋯ c 1 n c 21 c 22 ⋯ c 2 n ⋮ ⋮ ⋱ ⋮ c n 1 c n 2 ⋯ c n n ) = ( A 11 A 12 ⋯ A 1 n A 21 A 22 ⋯ A 2 n ⋮ ⋮ ⋱ ⋮ A n 1 A n 2 ⋯ A n n ) T = ( A 11 A 21 ⋯ A n 1 A 12 A 22 ⋯ A n 2 ⋮ ⋮ ⋱ ⋮ A 1 n A 2 n ⋯ A n n ) A^* =\begin{pmatrix} c_{11}& c_{12}& \cdots & c_{1n} \\ c_{21}& c_{22}& \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1}& c_{n2}& \cdots & c_{nn} \end{pmatrix} =\begin{pmatrix} A_{11}& A_{12}& \cdots & A_{1n} \\ A_{21}& A_{22}& \cdots & A_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1}& A_{n2}& \cdots & A_{nn} \end{pmatrix}^{\large{\rm{T}}} \\ =\begin{pmatrix} A_{11}& A_{21}& \cdots & A_{n1} \\ A_{12}& A_{22}& \cdots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n}& A_{2n}& \cdots & A_{nn} \end{pmatrix} A∗= c11c21⋮cn1c12c22⋮cn2⋯⋯⋱⋯c1nc2n⋮cnn = A11A21⋮An1A12A22⋮An2⋯⋯⋱⋯A1nA2n⋮Ann T= A11A12⋮A1nA21A22⋮A2n⋯⋯⋱⋯An1An2⋮Ann
b i j = ∑ k = 1 n a i k c k j = ∑ k = 1 n a i k A j k = { 0 i ≠ j ∣ A ∣ i = j b_{ij}=\sum_{k=1}^{n}a_{ik}c_{kj}=\sum_{k=1}^{n}a_{ik}A_{jk} =\begin{cases} 0&i\neq{j}\\ |A|&i=j \end{cases} bij=k=1∑naikckj=k=1∑naikAjk={0∣A∣i=ji=j
可见矩阵B是一个对角矩阵
∴ B = A A ∗ = ∣ A ∣ E \therefore{B=AA^*=|A|E} ∴B=AA∗=∣A∣E
类似的,可以证明 A ∗ A = ∣ A ∣ E A^*A=|A|E A∗A=∣A∣E
设 A A A可逆
( A − 1 ) − 1 = A (A^{-1})^{-1}=A (A−1)−1=A
( λ A ) − 1 = λ − 1 A − 1 (\lambda{A})^{-1}={\lambda^{-1}}A^{-1} (λA)−1=λ−1A−1
( A T ) − 1 = ( A − 1 ) T (A^T)^{-1}=(A^{-1})^T (AT)−1=(A−1)T
A , B A,B A,B可逆,则 A B AB AB可逆,且 ( A B ) − 1 = B − 1 A − 1 (AB)^{-1}=B^{-1}A^{-1} (AB)−1=B−1A−1
∵ ( A B ) ( B − 1 A − 1 ) = A ( B B − 1 ) A − 1 = A E A − 1 = A A − 1 = E \because{(AB)(B^{-1}A^{-1}})=A(BB^{-1})A^{-1}=AEA^{-1}=AA^{-1}=E ∵(AB)(B−1A−1)=A(BB−1)A−1=AEA−1=AA−1=E
∴ ( A B ) − 1 = B − 1 A − 1 \therefore{(AB)^{-1}=B^{-1}A^{-1}} ∴(AB)−1=B−1A−1
更一般的:
A A ∗ = A ∗ A = ∣ A ∣ E AA^*=A^*A=|A|E AA∗=A∗A=∣A∣E(数量阵)
∣ A − 1 ∣ = ∣ A ∣ − 1 |A^{-1}|=|A|^{-1} ∣A−1∣=∣A∣−1
( k A ) − 1 = k − 1 A − 1 (kA)^{-1}=k^{-1}A^{-1} (kA)−1=k−1A−1
( A B ) − 1 = B − 1 A − 1 (AB)^{-1}=B^{-1}A^{-1} (AB)−1=B−1A−1
∣ A ∗ ∣ = ∣ A ∣ n − 1 |A^*|=|A|^{n-1} ∣A∗∣=∣A∣n−1
( A ∗ ) − 1 = ( A − 1 ) ∗ = 1 ∣ A ∣ A , ( ∣ A ∣ ≠ 0 ) (A^*)^{-1}=(A^{-1})^*=\frac{1}{|A|}A,(|A|\neq{0}) (A∗)−1=(A−1)∗=∣A∣1A,(∣A∣=0)
因此,由 ∣ A ∣ ≠ 0 |A|\neq{0} ∣A∣=0可知: ∣ A ∗ ∣ ≠ 0 |A^*|\neq{0} ∣A∗∣=0,因为 ∣ A ∗ ∣ = ∣ A ∣ n − 1 ≠ 0 |A^*|=|A|^{n-1}\neq{0} ∣A∗∣=∣A∣n−1=0,即 A ∗ A^* A∗可逆
方法1:
由于 k E kE kE是可逆矩阵 ( k ≠ 0 ) (k\neq{0}) (k=0),所以 A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E两边都是可逆矩阵( k = ∣ A ∣ k=|A| k=∣A∣)
由 A A A可逆,可知B= A − 1 A^{-1} A−1是可逆的,且 A − 1 = A A^{-1}=A A−1=A
B B ∗ = ∣ B ∣ E BB^*=|B|E BB∗=∣B∣E,同时左乘 B − 1 B^{-1} B−1, B ∗ = ∣ B ∣ B − 1 B^*=|B|B^{-1} B∗=∣B∣B−1即 ( A − 1 ) ∗ = ∣ A − 1 ∣ ( A − 1 ) − 1 (A^{-1})^*=|A^{-1}|(A^{-1})^{-1} (A−1)∗=∣A−1∣(A−1)−1
而前面讨论过 ∣ A − 1 ∣ = ∣ A ∣ − 1 |A^{-1}|=|A|^{-1} ∣A−1∣=∣A∣−1,从而 ( A − 1 ) ∗ = ∣ A ∣ − 1 A (A^{-1})^*=|A|^{-1}A (A−1)∗=∣A∣−1A
可见 ( A ∗ ) − 1 = ( A − 1 ) ∗ = 1 ∣ A ∣ A , ( ∣ A ∣ ≠ 0 ) (A^*)^{-1}=(A^{-1})^*=\frac{1}{|A|}A,(|A|\neq{0}) (A∗)−1=(A−1)∗=∣A∣1A,(∣A∣=0)
方法2:
( k A ) ∗ = k n − 1 A ∗ (kA)^*=k^{n-1}A^* (kA)∗=kn−1A∗
( A ∗ ) T = ( A T ) ∗ (A^*)^T=(A^T)^* (A∗)T=(AT)∗
( A ∗ ) ∗ = ∣ A ∣ n − 2 A ( n ⩾ 2 ) (A^*)^*=|A|^{n-2}A(n\geqslant{2}) (A∗)∗=∣A∣n−2A(n⩾2)
综合运用上面得到的结论可以推导出来
记 B = A ∗ ; B ∗ = ∣ B ∣ B − 1 ; ( A ∗ ) ∗ = ∣ A ∗ ∣ ( A ∗ ) − 1 = ∣ A ∣ n − 1 ( ∣ A ∣ − 1 A ) = ∣ A ∣ n − 2 A 记B=A^*;B^*=|B|B^{-1};(A^*)^*=|A^*|(A^*)^{-1}=|A|^{n-1}(|A|^{-1}A)=|A|^{n-2}A 记B=A∗;B∗=∣B∣B−1;(A∗)∗=∣A∗∣(A∗)−1=∣A∣n−1(∣A∣−1A)=∣A∣n−2A