給定 $n \times n$ 的非奇異方陣 $A$，則下列性質成立
Claim: $adj A$ 為非奇異矩陣
Proof: 由於 $A$ 為 nonsingular，我們有
\[
A^{-1} = \frac{1}{detA} adj (A) \]
注意到等號左方 $A^{-1}$ 亦為 nonsingular (why? 由 $A$ 為 nonsingular 的定義出發，我們可知$A A^{-1} = I$ 等價說明 $A^{-1}$ 為 nonsingular)，且由於 $det A$ 僅為常數，故 $adj A$ 必定為 nonsingular。 $\square$
Claim: $\det (adj A) = (\det A)^{n-1}$
Proof: 由 $A^{-1}$ 可知
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow \left( {\det A} \right)\left( {{A^{ - 1}}} \right) = adj\left( A \right)\\
\Rightarrow \det \left( {\left( {\det A} \right)\left( {{A^{ - 1}}} \right)} \right) = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^n}\det \left( {{A^{ - 1}}} \right) = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^n}\frac{1}{{\det A}} = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^{n - 1}} = \det \left( {adj\left( A \right)} \right)
\end{array}\]
Claim: $(adj A)^{-1} = adj (A^{-1}) = \frac{1}{det A} A$
首先注意到
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow \left( {\det A} \right)I = adj\left( A \right)A
\end{array}\]由 Claim 1 可知 $adj A$ 為 nonsingular 故 $(adj (A))^{-1}$ 存在，亦即我們可改寫上式如下
\[\begin{array}{l}
{\left( {adj\left( A \right)} \right)^{ - 1}}\left( {\det A} \right)I = {\left( {adj\left( A \right)} \right)^{ - 1}}adj\left( A \right)A\\
\Rightarrow \left( {\det A} \right){\left( {adj\left( A \right)} \right)^{ - 1}} = A\\
\Rightarrow {\left( {adj\left( A \right)} \right)^{ - 1}} = \frac{1}{{\det A}}A
\end{array}\]
接著我們證明 $adj (A^{-1}) = \frac{1}{det A} A$
我們觀察
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow A = \frac{1}{{\det \left( {{A^{ - 1}}} \right)}}adj\left( {{A^{ - 1}}} \right)\\
\Rightarrow A = \left( {\det A} \right)adj\left( {{A^{ - 1}}} \right)\\
\Rightarrow \frac{1}{{\det A}}A = adj\left( {{A^{ - 1}}} \right)
\end{array}\]
故綜合以上所述，我們可得 $(adj A)^{-1} = adj (A^{-1}) = \frac{1}{det A} A$ $\square$
Claim: $adj A$ 為非奇異矩陣
Proof: 由於 $A$ 為 nonsingular，我們有
\[
A^{-1} = \frac{1}{detA} adj (A) \]
注意到等號左方 $A^{-1}$ 亦為 nonsingular (why? 由 $A$ 為 nonsingular 的定義出發，我們可知$A A^{-1} = I$ 等價說明 $A^{-1}$ 為 nonsingular)，且由於 $det A$ 僅為常數，故 $adj A$ 必定為 nonsingular。 $\square$
Claim: $\det (adj A) = (\det A)^{n-1}$
Proof: 由 $A^{-1}$ 可知
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow \left( {\det A} \right)\left( {{A^{ - 1}}} \right) = adj\left( A \right)\\
\Rightarrow \det \left( {\left( {\det A} \right)\left( {{A^{ - 1}}} \right)} \right) = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^n}\det \left( {{A^{ - 1}}} \right) = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^n}\frac{1}{{\det A}} = \det \left( {adj\left( A \right)} \right)\\
\Rightarrow {\left( {\det A} \right)^{n - 1}} = \det \left( {adj\left( A \right)} \right)
\end{array}\]
Claim: $(adj A)^{-1} = adj (A^{-1}) = \frac{1}{det A} A$
首先注意到
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow \left( {\det A} \right)I = adj\left( A \right)A
\end{array}\]由 Claim 1 可知 $adj A$ 為 nonsingular 故 $(adj (A))^{-1}$ 存在，亦即我們可改寫上式如下
\[\begin{array}{l}
{\left( {adj\left( A \right)} \right)^{ - 1}}\left( {\det A} \right)I = {\left( {adj\left( A \right)} \right)^{ - 1}}adj\left( A \right)A\\
\Rightarrow \left( {\det A} \right){\left( {adj\left( A \right)} \right)^{ - 1}} = A\\
\Rightarrow {\left( {adj\left( A \right)} \right)^{ - 1}} = \frac{1}{{\det A}}A
\end{array}\]
接著我們證明 $adj (A^{-1}) = \frac{1}{det A} A$
我們觀察
\[\begin{array}{l}
{A^{ - 1}} = \frac{1}{{\det A}}adj\left( A \right)\\
\Rightarrow A = \frac{1}{{\det \left( {{A^{ - 1}}} \right)}}adj\left( {{A^{ - 1}}} \right)\\
\Rightarrow A = \left( {\det A} \right)adj\left( {{A^{ - 1}}} \right)\\
\Rightarrow \frac{1}{{\det A}}A = adj\left( {{A^{ - 1}}} \right)
\end{array}\]
故綜合以上所述，我們可得 $(adj A)^{-1} = adj (A^{-1}) = \frac{1}{det A} A$ $\square$
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