# 余因子

$n$ 次の正方行列 $A$

$A=\left(\begin{array}{cccc}{a}_{11}& \cdots & \cdots & {a}_{1n}\\ ⋮& \mathrm{\ddots }& & ⋮\\ ⋮& & \mathrm{\ddots }& ⋮\\ {a}_{n1}& \cdots & \cdots & {a}_{nn}\end{array}\right)$

${a}_{ij}$余因子( ${\stackrel{˜}{a}}_{ij}$ と書く)を

${\stackrel{˜}{a}}_{ij}={\left(-1\right)}^{i+j}\left|\begin{array}{cccccc}{a}_{11}& \cdots & {a}_{1\left(j-1\right)}& {a}_{1\left(j+1\right)}& \cdots & {a}_{1n}\\ ⋮& & ⋮& ⋮& & \\ {a}_{\left(i-1\right)1}& \cdots & & & \cdots & {a}_{\left(i-1\right)1}\\ {a}_{\left(i+1\right)1}& \cdots & & & \cdots & {a}_{\left(i+1\right)1}\\ ⋮& & ⋮& ⋮& & \\ {a}_{n1}& \cdots & {a}_{n\left(j-1\right)}& {a}_{n\left(j+1\right)}& \cdots & {a}_{nn}\end{array}\right|$

と定める．つまり， $A$の行列式$|A|$ から $j$列の${a}_{1j}$${a}_{nj}$までの成分と，$i$行の${a}_{i1}$${a}_{in}$ 　までの成分を削除し得られる $\left(n-1\right)$ 次の行列式に

${\left(-1\right)}^{i+j}$ 　をかけたものを余因子という．

## ■具体例

$A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$

 ${\stackrel{˜}{a}}_{11}$ $={\left(-1\right)}^{1+1}\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|$ $=\left(5×9-8×6\right)$ $=-3$

 ${\stackrel{˜}{a}}_{12}$ $={\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|$ $=\left(4×9-7×6\right)$ $=-6$

 ${\stackrel{˜}{a}}_{13}$ $={\left(-1\right)}^{1+3}\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|$ $=\left(4×8-7×5\right)$ $=-3$

 ${\stackrel{˜}{a}}_{21}$ $={\left(-1\right)}^{2+1}\left|\begin{array}{cc}2& 3\\ 8& 9\end{array}\right|$ $=\left(2×9-8×3\right)$ $=-6$

 ${\stackrel{˜}{a}}_{22}$ $={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 3\\ 7& 9\end{array}\right|$ $= 1 × 9 − 7 × 3$ $=-12$

 ${\stackrel{˜}{a}}_{23}$ $={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|$ $=\left(1×8-7×2\right)$ $=-6$

 ${\stackrel{˜}{a}}_{31}$ $={\left(-1\right)}^{3+1}\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|$ $=\left(2×6-5×3\right)$ $=-3$

 ${\stackrel{˜}{a}}_{32}$ $={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|$ $=\left(1×6-4×3\right)$ $=-6$

 ${\stackrel{˜}{a}}_{33}$ $={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|$ $=\left(1×5-4×2\right)$ $=-3$

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