关键公式：
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|AB|=|A||B|
∣AB∣=∣A∣∣B∣
#伴随矩阵及其运算
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(1)对于任意
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n阶方阵
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A^*
A∗,且有
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AA^*=A^*A=|A|E,|A^*|=|A|^{n-1}
AA∗=A∗A=∣A∣E,∣A∗∣=∣A∣n−1
按照定义：
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\begin{aligned}&A^{-1}=\dfrac{A^*}{|A|}\\ &所以 AA^*=A^*A=|A|E\\ &A^*=|A|A^{-1}\\ &|A^*|=\left||A|A^{-1}\right|\\ &=|A|^{n-1}\end{aligned}
A−1=∣A∣A∗所以AA∗=A∗A=∣A∣EA∗=∣A∣A−1∣A∗∣=∣∣∣A∣A−1∣∣=∣A∣n−1
当
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∣A∤=0时，有
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\begin{aligned}&A^*=|A|A^{-1},A^{-1}=\dfrac{1}{|A|}A^*,A=|A|(A^*)^{-1}\\ &(kA)(kA)^*=|kA|E\\ &A^T(A^T)^*=|A^T|E\\ &A^{-1}(A^{-1})^*=|A^{-1}|E\\ &A^*(A^*)^*=|A^*|E\\ \end{aligned}
A∗=∣A∣A−1,A−1=∣A∣1A∗,A=∣A∣(A∗)−1(kA)(kA)∗=∣kA∣EAT(AT)∗=∣AT∣EA−1(A−1)∗=∣A−1∣EA∗(A∗)∗=∣A∗∣E
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(A^T)^*=(A^*)^T,(A^{-1})^*=(A^*)^{-1},(AB)^*=B^*A^*,(A^*)^*=|A|^{n-2}A
(AT)∗=(A∗)T,(A−1)∗=(A∗)−1,(AB)∗=B∗A∗,(A∗)∗=∣A∣n−2A
对于
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A=\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}
A=⎝⎜⎜⎜⎛a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎠⎟⎟⎟⎞
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A^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}
AT=⎝⎜⎜⎜⎛a11a12⋮a1na21a22⋮a2n⋯⋯⋱⋯an1an2⋮ann⎠⎟⎟⎟⎞
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A^*=|A|A^{-1},A^{-1}=\dfrac{1}{|A|}A^*,A=|A|(A^*)^{-1}
A∗=∣A∣A−1,A−1=∣A∣1A∗,A=∣A∣(A∗)−1
这个不需要证明：
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(kA)(kA)^*=|kA|E
(kA)(kA)∗=∣kA∣E
这个显然
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A^T(A^T)^*=|A^T|E
AT(AT)∗=∣AT∣E
这个显然
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A^{-1}(A^{-1})^*=|A^{-1}|E
A−1(A−1)∗=∣A−1∣E
这个显然
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A^*(A^*)^*=|A^*|E
A∗(A∗)∗=∣A∗∣E
这个显然
这些都只需要验证对应的矩阵是否可逆，
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|A^*|=\left||A|A^{-1}\right|=|A|^{n-1}\neq0
∣A∗∣=∣∣∣A∣A−1∣∣=∣A∣n−1̸=0
∣
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|A^T|=|A|\neq0
∣AT∣=∣A∤=0
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|A^{-1}|=|A|^{-1}\neq0
∣A−1∣=∣A∣−1̸=0
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|kA|=k^n|A|\neq0
∣kA∣=kn∣A∤=0
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(A^T)^*=(A^*)^T
(AT)∗=(A∗)T
对于
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A=\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}
A=⎝⎜⎜⎜⎛a11a21⋮an1a12a22⋮an2⋯⋯⋱⋯a1na2n⋮ann⎠⎟⎟⎟⎞
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A^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}
AT=⎝⎜⎜⎜⎛a11a12⋮a1na21a22⋮a2n⋯⋯⋱⋯an1an2⋮ann⎠⎟⎟⎟⎞
显然
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(A^{-1})^*=(A^*)^{-1}
(A−1)∗=(A∗)−1
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(A^*)^{-1}=(|A|A^{-1})^{-1}=\dfrac{1}{|A|}A\\ (A^{-1})^*=|A^{-1}|(A^{-1})^{-1}=\dfrac{1}{|A|}A=(A^*)^{-1}
(A∗)−1=(∣A∣A−1)−1=∣A∣1A(A−1)∗=∣A−1∣(A−1)−1=∣A∣1A=(A∗)−1
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(AB)^*=B^*A^*
(AB)∗=B∗A∗
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(AB)^*=|AB|(AB)^{-1}=|A||B|B^{-1}A^{-1}\\ B^*A^*=|B|B^{-1}|A|A^{-1}
(AB)∗=∣AB∣(AB)−1=∣A∣∣B∣B−1A−1B∗A∗=∣B∣B−1∣A∣A−1
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(A^*)^*=|A|^{n-2}A
(A∗)∗=∣A∣n−2A
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A^*(A^*)^*=|A^*|E\\ (A^*)^*=|A^*|(A^*)^{-1}E\\ =|A|^{n-1}(|A|A^{-1})^{-1}\\ =|A|^{n-2}A
A∗(A∗)∗=∣A∗∣E(A∗)∗=∣A∗∣(A∗)−1E=∣A∣n−1(∣A∣A−1)−1=∣A∣n−2A
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