LA_矩阵运算的性质@方阵取行列式@取伴随@取逆@转置
- 矩阵运算的性质@方阵取行列式@取伴随@取逆@转置
可逆矩阵@矩阵的逆🎈
- 矩阵和实数相仿,具有加/减/乘三种运算
- 数的乘法的逆运算是除法,相对应矩阵乘法的逆运算用矩阵的逆来描述
- 数a的倒数: a − 1 = 1 a a^{-1}=\frac{1}{a} a−1=a1, a a − 1 = 1 aa^{-1}=1 aa−1=1
- 可逆矩阵A的逆: A − 1 满足 A A − 1 = E A^{-1}满足AA^{-1}=E A−1满足AA−1=E
- 若A可逆,则A的逆矩阵唯一
- A可逆 ⇔ ∣ A ∣ ≠ 0 \Leftrightarrow|A|\neq{0} ⇔∣A∣=0
- 设 A , B A,B A,B是n阶矩阵且 A B = E , 则 B A = E AB=E,则BA=E AB=E,则BA=E(A,B互为逆矩阵)
n阶矩阵A可逆的充要条件
- 存在n阶矩阵B,使得 A B = E 或 B A = E AB=E或BA=E AB=E或BA=E
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∣
A
∣
≠
0
|A|\neq{0}
∣A∣=0,即A是非奇异的(由:Cramer’s Rule及其推论有):
- 齐次方程组 A x = 0 Ax=\bold0 Ax=0只有唯一解(零解)
- ∀ b \forall{b} ∀b,非齐次线性方程组 A x = b Ax=b Ax=b总有唯一解(非零解)
- 秩 r ( A ) = n 秩r(A)=n 秩r(A)=n
- A的行(列)向量线性无关
- 矩阵A的特征值不全为0
- 或者说:A的特征方程 ∣ λ E − A ∣ = 0 |\lambda{E}-A|=0 ∣λE−A∣=0的全部n个根: λ i , ( i = 1 , 2 , ⋯ , n ) \lambda_i,(i=1,2,\cdots,n) λi,(i=1,2,⋯,n), ∏ i = 1 n λ i ≠ 0 \prod_{i=1}^{n}\limits\lambda_{i}\neq{0} i=1∏nλi=0
- A可以表示成一些初等矩阵的乘积: ∏ i = 1 P i A = E \prod\limits_{i=1}P_iA=E i=1∏PiA=E
A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E的证明
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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
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设 A = ( a i j ) A=(a_{ij}) A=(aij), B = A A ∗ = ( b i j ) B=AA^*=(b_{ij}) B=AA∗=(bij)
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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
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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
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可见矩阵B是一个对角矩阵
- B = ( b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n n ) = ( ∣ A ∣ 0 ∣ A ∣ ⋱ 0 ∣ A ∣ ) = ∣ A ∣ E B=\begin{pmatrix} b_{11}& b_{12}& \cdots & b_{1n} \\ b_{21}& b_{22}& \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{nn} \end{pmatrix} =\begin{pmatrix} |A|& & & \huge{0} \\ & |A|& & \\ & & \ddots & \\ \huge{0} & & & |A| \end{pmatrix} =|A|E B= b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1nb2n⋮bnn = ∣A∣0∣A∣⋱0∣A∣ =∣A∣E
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∴ B = A A ∗ = ∣ A ∣ E \therefore{B=AA^*=|A|E} ∴B=AA∗=∣A∣E
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类似的,可以证明 A ∗ A = ∣ A ∣ E A^*A=|A|E A∗A=∣A∣E
可逆矩阵的性质🎈
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设 A A A可逆
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( A − 1 ) − 1 = A (A^{-1})^{-1}=A (A−1)−1=A
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( λ A ) − 1 = λ − 1 A − 1 (\lambda{A})^{-1}={\lambda^{-1}}A^{-1} (λA)−1=λ−1A−1
- 容易验证: ( λ A ) ( λ − 1 A − 1 ) = E (\lambda{A})(\lambda^{-1}A^{-1})=E (λA)(λ−1A−1)=E,所以 λ A \lambda{A} λA可逆,且 ( λ A ) − 1 = λ − 1 A − 1 (\lambda{A})^{-1}=\lambda^{-1}A^{-1} (λA)−1=λ−1A−1
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( A T ) − 1 = ( A − 1 ) T (A^T)^{-1}=(A^{-1})^T (AT)−1=(A−1)T
- 由矩阵乘法的转置性质: ( A B ) T = B T A T (AB)^T=B^TA^T (AB)T=BTAT,即 A T B T = ( B A ) T A^TB^T=(BA)^T ATBT=(BA)T
- 方法1:
- 等号左边是 A T A^T AT的逆, A T ( A T ) − 1 = E A^T(A^T)^{-1}=E AT(AT)−1=E
- 欲证上式,等价于证等号右边也是 A T A^T AT的逆: A T ( A − 1 ) T = ( A − 1 A ) T = E A^T(A^{-1})^T=(A^{-1}A)^T=E AT(A−1)T=(A−1A)T=E
- 可见等号右边也是 A T A^T AT的逆
- 从而 ( A T ) − 1 = ( A − 1 ) T (A^T)^{-1}=(A^{-1})^{T} (AT)−1=(A−1)T
- 方法2(使用换元法描述):
- 令 B = ( A T ) − 1 B=(A^T)^{-1} B=(AT)−1
- C = ( A − 1 ) T C=(A^{-1})^T C=(A−1)T
- A T B = A T ( A T ) − 1 = E A^TB=A^T(A^T)^{-1}=E ATB=AT(AT)−1=E
- A T C = ( A T ) ( A − 1 ) T = ( A − 1 A ) T = E A^TC=(A^T)(A^{-1})^T=(A^{-1}A)^T=E ATC=(AT)(A−1)T=(A−1A)T=E
- 因此B,C都是 A T A^T AT的逆,所以 B = C = ( A T ) − 1 B=C=(A^T)^{-1} B=C=(AT)−1
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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
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∵ ( 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
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∴ ( A B ) − 1 = B − 1 A − 1 \therefore{(AB)^{-1}=B^{-1}A^{-1}} ∴(AB)−1=B−1A−1
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更一般的:
- ( A 1 A 2 ⋯ A m ) − 1 = A m − 1 A m − 1 − 1 ⋯ A 1 − 1 (A_1A_2\cdots{A_m})^{-1}=A_m^{-1}A_{m-1}^{-1}\cdots{A_1^{-1}} (A1A2⋯Am)−1=Am−1Am−1−1⋯A1−1
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伴随运算@伴随矩阵@逆矩阵的公式🎈
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A A ∗ = A ∗ A = ∣ A ∣ E AA^*=A^*A=|A|E AA∗=A∗A=∣A∣E(数量阵)
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∣
A
∣
−
1
A
A
∗
=
∣
A
∣
−
1
A
∗
A
=
E
|A|^{-1}AA^*=|A|^{-1}A^*A=E
∣A∣−1AA∗=∣A∣−1A∗A=E
- ∣ A − 1 ∣ A A ∗ = A ∣ A − 1 ∣ A ∗ |A^{-1}|AA^*=A|A^{-1}|A^* ∣A−1∣AA∗=A∣A−1∣A∗
- A − 1 = ∣ A ∣ − 1 A ∗ A^{-1}=|A|^{-1}A^* A−1=∣A∣−1A∗
-
A
∗
=
∣
A
∣
A
−
1
A^*=|A|A^{-1}
A∗=∣A∣A−1(A可逆(不是所有方阵的伴随矩阵都可以展开成右侧))
- 对 A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E两边左乘 A − 1 A^{-1} A−1得到.
- 主要用于推导其他关于伴随矩阵的结论)
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∣
A
∣
−
1
A
A
∗
=
∣
A
∣
−
1
A
∗
A
=
E
|A|^{-1}AA^*=|A|^{-1}A^*A=E
∣A∣−1AA∗=∣A∣−1A∗A=E
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∣ A − 1 ∣ = ∣ A ∣ − 1 |A^{-1}|=|A|^{-1} ∣A−1∣=∣A∣−1
- A A − 1 = E AA^{-1}=E AA−1=E,对两边取行列式
- ∣ A A − 1 ∣ = ∣ E ∣ |AA^{-1}|=|E| ∣AA−1∣=∣E∣,即 ∣ A ∣ ∣ A − 1 ∣ = 1 |A||A^{-1}|=1 ∣A∣∣A−1∣=1
- 从而 ∣ A − 1 ∣ = ∣ A ∣ − 1 |A^{-1}|=|A|^{-1} ∣A−1∣=∣A∣−1
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( k A ) − 1 = k − 1 A − 1 (kA)^{-1}=k^{-1}A^{-1} (kA)−1=k−1A−1
- B = k A B=kA B=kA
- C = k − 1 A − 1 C=k^{-1}A^{-1} C=k−1A−1
- B C = ( k A ) ( k − 1 A − 1 ) = k k − 1 A A − 1 = E BC=(kA)(k^{-1}A^{-1})=kk^{-1}AA^{-1}=E BC=(kA)(k−1A−1)=kk−1AA−1=E
- ∴ B − 1 = C , ( k A ) − 1 = k − 1 A − 1 \therefore{B^{-1}=C},(kA)^{-1}=k^{-1}A^{-1} ∴B−1=C,(kA)−1=k−1A−1
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( A B ) − 1 = B − 1 A − 1 (AB)^{-1}=B^{-1}A^{-1} (AB)−1=B−1A−1
- 令 C = A B C=AB C=AB, D = B − 1 A − 1 D=B^{-1}A^{-1} D=B−1A−1
- C D = A B B − 1 A − 1 = A ( B B − 1 ) A − 1 = A E A − 1 = A A − 1 = E CD=ABB^{-1}A^{-1}=A(BB^{-1})A^{-1}=AEA^{-1}=AA^{-1}=E CD=ABB−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
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∣ A ∗ ∣ = ∣ A ∣ n − 1 |A^*|=|A|^{n-1} ∣A∗∣=∣A∣n−1
- 对 A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E两边取行列式, ∣ A ∣ ∣ A ∗ ∣ = ∣ A ∣ n ∣ E ∣ = ∣ A ∣ n |A||A^*|=|A|^n|E|=|A|^n ∣A∣∣A∗∣=∣A∣n∣E∣=∣A∣n,所以: ∣ A ∗ ∣ = ∣ A ∣ n − 1 |A^*|=|A|^{n-1} ∣A∗∣=∣A∣n−1
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( 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)
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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
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方法1:
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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∣)
- A A ∗ = ∣ A ∣ E AA^*=|A|E AA∗=∣A∣E的两边同时取逆,得到 ( A A ∗ ) − 1 = ( ∣ A ∣ E ) − 1 (AA^*)^{-1}=(|A|E)^{-1} (AA∗)−1=(∣A∣E)−1,即 ( A ∗ ) − 1 A − 1 = ∣ A ∣ − 1 E − 1 = ∣ A ∣ − 1 E (A^{*})^{-1}A^{-1}=|A|^{-1}E^{-1}=|A|^{-1}E (A∗)−1A−1=∣A∣−1E−1=∣A∣−1E
- 对 ( A ∗ ) − 1 A − 1 = ∣ A ∣ − 1 E (A^*)^{-1}A^{-1}=|A|^{-1}E (A∗)−1A−1=∣A∣−1E两边同时右乘 A A A,得: ( A ∗ ) − 1 = ∣ A ∣ − 1 A (A^*)^{-1}=|A|^{-1}A (A∗)−1=∣A∣−1A
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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
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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
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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
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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)
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方法2:
- 对
A
∗
=
∣
A
∣
A
−
1
A^*=|A|A^{-1}
A∗=∣A∣A−1两边取逆:
(
A
∗
)
−
1
=
∣
A
∣
−
1
A
(A^*)^{-1}=|A|^{-1}A
(A∗)−1=∣A∣−1A
- 令 B = A − 1 B=A^{-1} B=A−1,由于A可逆,B也可逆
- B ∗ = ∣ B ∣ B − 1 = ∣ A − 1 ∣ A = ∣ A ∣ − 1 A B^*=|B|B^{-1}=|A^{-1}|A=|A|^{-1}A B∗=∣B∣B−1=∣A−1∣A=∣A∣−1A
- 所以 ( A ∗ ) − 1 = ( A − 1 ) ∗ = ∣ A ∣ − 1 A (A^*)^{-1}=(A^{-1})^*=|A|^{-1}A (A∗)−1=(A−1)∗=∣A∣−1A
- 对
A
∗
=
∣
A
∣
A
−
1
A^*=|A|A^{-1}
A∗=∣A∣A−1两边取逆:
(
A
∗
)
−
1
=
∣
A
∣
−
1
A
(A^*)^{-1}=|A|^{-1}A
(A∗)−1=∣A∣−1A
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( k A ) ∗ = k n − 1 A ∗ (kA)^*=k^{n-1}A^* (kA)∗=kn−1A∗
- 使用换元法并结合基础性质展开来推导
- 由于A可逆, k A kA kA也可逆
- 令 B = k A B=kA B=kA,则B可逆, B B ∗ = ∣ B ∣ E , B ∗ = ∣ B ∣ B − 1 , ( k A ) ∗ = ∣ k A ∣ ( k A ) − 1 = k n ∣ A ∣ k − 1 A − 1 BB^*=|B|E,B^*=|B|B^{-1},(kA)^*=|kA|(kA)^{-1}=k^n|A|k^{-1}A^{-1} BB∗=∣B∣E,B∗=∣B∣B−1,(kA)∗=∣kA∣(kA)−1=kn∣A∣k−1A−1
- 所以 ( k A ) ∗ = k n − 1 ∣ A ∣ A − 1 = k n − 1 A ∗ (kA)^*=k^{n-1}|A|A^{-1}=k^{n-1}A^* (kA)∗=kn−1∣A∣A−1=kn−1A∗
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( A ∗ ) T = ( A T ) ∗ (A^*)^T=(A^T)^* (A∗)T=(AT)∗
- 借助转置的性质推导
- A ∗ = ( ∣ A ∣ A − 1 ) , ( A ∗ ) T = ( ∣ A ∣ A − 1 ) T = ∣ A ∣ ( A − 1 ) T A^*=(|A|A^{-1}),(A^*)^T=(|A|A^{-1})^{T}=|A|(A^{-1})^T A∗=(∣A∣A−1),(A∗)T=(∣A∣A−1)T=∣A∣(A−1)T
- 记 B = A T , B ∗ = ∣ B ∣ B − 1 , 即 ( A T ) ∗ = ∣ A T ∣ ( A T ) − 1 = ∣ A ∣ ( A − 1 ) T 记B=A^T,B^*=|B|B^{-1},即(A^T)^*=|A^T|(A^T)^{-1}=|A|(A^{-1})^{T} 记B=AT,B∗=∣B∣B−1,即(AT)∗=∣AT∣(AT)−1=∣A∣(A−1)T
- 所以 ( A ∗ ) T = ( A T ) ∗ = ∣ A ∣ ( A T ) − 1 (A^*)^T=(A^T)^*=|A|(A^T)^{-1} (A∗)T=(AT)∗=∣A∣(AT)−1
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( A ∗ ) ∗ = ∣ A ∣ n − 2 A ( n ⩾ 2 ) (A^*)^*=|A|^{n-2}A(n\geqslant{2}) (A∗)∗=∣A∣n−2A(n⩾2)
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综合运用上面得到的结论可以推导出来
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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
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