正則上三角行列の逆行列公式

前提

$n$ を正の整数とし，$E_{n}$ を $n$ 次単位行列とする。

以下では，$n$ 次複素行列$A \in \mathbb{C}^{n \times n}$ を，正則な上三角行列とし，
\begin{align} \qquad \begin{aligned} A = \left[ \begin{array}{ccc} a_{1,1} & \cdots & a_{1,n} \\ {} & \ddots & \vdots \\ {\Huge\boldsymbol{0}} & & a_{n,n} \\ \end{array} \right] \end{aligned} \end{align}
と表す。このとき，
\begin{align} \qquad \begin{aligned} \left| A \right| = \det A = a_{1,1} \, \cdots \, a_{n,n} = \prod_{r=1}^{n} a_{r,r} \qquad \textsf{かつ} \qquad \left| A \right| \neq 0 \end{aligned} \end{align}
より，
\begin{align} \qquad \begin{aligned} a_{r,r}^{} \neq 0 \qquad \left( \, \boldsymbol{\forall} \, r \in \left\{ 1 ,\, \dots ,\, n \right\} \, \right) \end{aligned} \end{align}
であることに注意する。
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$A$ の逆行列を求めるには通常，掃き出し法（ガウスの消去法）等を用いるのが一般的ですが，ここでは一般的な手法は使わず，また計算量も度外視して，$A^{-1}$ の systematic な作り方を解説していきます。

とにかく丁寧に解説した（つもりな）ので，手加減なしの行列式計算がヤバめになっています。
（行列式計算アレルギーをお持ちの方は閲覧注意！）

なお，$n = 1 ,\, 2 ,\, 3,\, 4 ,\, 5$ の場合の$A$ の成分表示については，分かりやすさのために，添え字を簡略化する方向で調節しています。
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$n = 1$ の場合

$\displaystyle A = \left[ \begin{array}{c} a \end{array} \right] = a $ と表す。このとき，$ \displaystyle A^{-1} = \left[ \begin{array}{c} a^{-1} \end{array} \right] = a^{-1} $ である。
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$n = 2$ の場合

$\displaystyle A = \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] $ と表す。このとき，$ \displaystyle A' = \left[ \begin{array}{c|c} \cellcolor{ pink } a'_{1} & \cellcolor{ lightgreen } b' \\ \hline 0 & \cellcolor{ pink } a'_{2} \\ \end{array} \right] $ が $A$ の逆行列となるように成分を決めていく。
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1. $\textsf{[ 対角成分 $a'_{1} \, \boldsymbol{,} \ a'_{2}$ の決め方 ]}$

   ${}$
   \begin{align} \begin{aligned} \textsf{(1-a)} \quad & \textsf{$A'$ の $( 1 \boldsymbol{,} \, 1 )$ 成分} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a'_{1} & b' \\ \hline 0 & a'_{2} \\ \end{array} \right] \quad \textsf{を決めるには，$A$ の $( 1 \boldsymbol{,} \, 1 )$ 成分} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & b \\ \hline 0 & a_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & \begin{aligned} a'_{1} = a_{1}^{-1} \end{aligned} \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(1-b)} \quad & \textsf{$A'$ の $( 2 \boldsymbol{,} \, 2 )$ 成分} \quad \left[ \begin{array}{c|c} a'_{1} & b' \\ \hline 0 & \cellcolor{ pink } a'_{2} \\ \end{array} \right] \quad \textsf{も同様に，$A$ の $( 2 \boldsymbol{,} \, 2 )$ 成分} \quad \left[ \begin{array}{c|c} a_{1} & b \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & \begin{aligned} a'_{2} = a_{2}^{-1} \end{aligned} \\[10pt] & \textsf{と決める。} \end{aligned} \end{align}
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2. $\textsf{[ $( 1 \boldsymbol{,} \, 2 )$ 成分 $b'$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 2 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c} a'_{1} & \cellcolor{ lightgreen } b' \\ \hline 0 & a'_{2} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(2-a)} \quad & \textsf{$A$ の $( 1 \boldsymbol{,} \, 2 )$ 成分} \quad \left[ \begin{array}{c|c} a_{1} & \cellcolor{ lightgreen } b \\ \hline 0 & a_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & b' = (-1)^{1+2} \cdot \dfrac{ b }{ \fbox{$\phantom{\det A}$} } = - \dfrac{b}{ \fbox{$\phantom{\det A}$} } \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
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   \begin{align} \begin{aligned} \textsf{(2-b)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & b \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & b' = - \dfrac{b}{ \det A } = - \dfrac{b}{ a_{1} a_{2} } \\[10pt] & \textsf{として分母を決める。} \end{aligned} \end{align}
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3. $\textsf{[ 成分計算 ]}$

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   ゆえに，
   \begin{align} \qquad \begin{aligned} A A' &= \left[ \begin{array}{cc} a_{1} & b \\ 0 & a_{2} \\ \end{array} \right] \left[ \begin{array}{cc} a'_{1} & b' \\ 0 & a'_{2} \\ \end{array} \right] = \left[ \begin{array}{c|c} a_{1} a'_{1} & a_{1} b' + b a'_{2} \\[5pt] \hline 0 & a_{1} a'_{2} \\ \end{array} \right] \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad & \begin{aligned} a_{1} a'_{1} = a_{2} a'_{2} = 1 \, \boldsymbol{,} \end{aligned} \\[10pt] & \begin{aligned} a_{1} b' + b a'_{2} &= a_{1} b' + \dfrac{b}{a_{2}} = a_{1} b' - a_{1} \cdot \left( - \dfrac{b}{ a_{1} a_{2} } \right) = a_{1} b' - a_{1} b' = 0 \, \boldsymbol{.} \end{aligned} \end{align}
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4. $\textsf{[ 計算終了 ]}$

   ${}$
   よって，$A A' = E_{2}$ となるので，$A' = A^{-1}$ である。
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$n = 3$ の場合

$\displaystyle A = \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline 0 & \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] $ と表す。このとき，$ \displaystyle A' = \left[ \begin{array}{c|c|c} \cellcolor{ pink } a'_{1} & \cellcolor{ lightgreen } b'_{1} & \cellcolor{ lightsalmon } c' \\ \hline 0 & \cellcolor{ pink } a'_{2} & \cellcolor{ lightgreen } b'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} \\ \end{array} \right] $ が $A$ の逆行列となるように成分を決めていく。
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1. $\textsf{[ 対角成分 $a'_{1} \, \boldsymbol{,} \ a'_{2} \, \boldsymbol{,} \ a'_{3}$ の決め方 ]}$

   ${}$
   $A'$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c} \cellcolor{ pink } a'_{1} & b'_{1} & c' \\ \hline 0 & \cellcolor{ pink } a'_{2} & b'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} \\ \end{array} \right] $ を決めるには，$A$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] $ のそれぞれに着目し，
   \begin{align} \qquad \begin{aligned} a'_{1} = a_{1}^{-1} \, \boldsymbol{,} \qquad a'_{2} = a_{2}^{-1} \, \boldsymbol{,} \qquad a'_{3} = a_{3}^{-1} \end{aligned} \end{align}
   と決める。
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2. $\textsf{[ 対角隣接成分 $b'_{1} \, \boldsymbol{,} \, b'_{2}$ の決め方 ]}$

   ${}$
   $A'$ の対角隣接成分 $ \displaystyle \left[ \begin{array}{c|c|c} a'_{1} & \cellcolor{ lightgreen } b'_{1} & c' \\ \hline 0 & a'_{2} & \cellcolor{ lightgreen } b'_{2} \\ \hline 0 & 0 & a'_{3} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(2-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] \quad \textsf{に着目し，$n = 2$ の場合と同様に，} \quad \\[10pt] & b'_{1} = (-1)^{1+2} \cdot \dfrac{ b_{1} }{ a_{1} a_{2} } = - \dfrac{ b_{1} }{ a_{1} a_{2} } \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
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   \begin{align} \begin{aligned} \textsf{(2-b)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \quad \textsf{に着目し，$n = 2$ の場合と同様に，} \quad \\[10pt] & b'_{2} = (-1)^{2+3} \cdot \dfrac{ b_{2} }{ a_{2} a_{3} } = - \dfrac{ b_{2} }{ a_{2} a_{3} } \\[10pt] & \textsf{と決める。} \end{aligned} \end{align}
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3. $\textsf{[ $( 1 \boldsymbol{,} \, 3 )$ 成分 $c'$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 3 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c} a'_{1} & b'_{1} & \cellcolor{ lightsalmon } c' \\ \hline 0 & a'_{2} & b'_{2} \\ \hline 0 & 0 & a'_{3} \\ \end{array} \right] $ を決めるには，
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   \begin{align} \begin{aligned} \textsf{(3-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c' = (-1)^{1+3} \cdot \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| = \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-b)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c' = \dfrac{ 1 }{ \det A } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| = \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| \\[10pt] & \textsf{として分母を決める。} \end{aligned} \end{align}
   ${}$

4. $\textsf{[ 成分計算 ]}$

   ${}$
   ゆえに，
   \begin{align} \qquad \begin{aligned} A A' &= \left[ \begin{array}{ccc} a_{1} & b_{1} & c \\ 0 & a_{2} & b_{2} \\ 0 & 0 & a_{3} \\ \end{array} \right] \left[ \begin{array}{ccc} a'_{1} & b'_{1} & c' \\ 0 & a'_{2} & b'_{2} \\ 0 & 0 & a'_{3} \\ \end{array} \right] \\[10pt] &= \left[ \begin{array}{c|c|c} a_{1} a'_{1} & a_{1} b'_{1} + b_{1} a'_{2} & a_{1} c' + b_{1} b'_{2} + c a'_{3} \\[5pt] \hline 0 & a_{2} a'_{2} & a_{2} b'_{2} + b_{2} a'_{3} \\[5pt] \hline 0 & 0 & a_{3} a'_{3} \\ \end{array} \right] \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad & \begin{aligned} a_{1} a'_{1} = a_{2} a'_{2} = a_{3} a'_{3} = 1 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} b'_{1} + b_{1} a'_{2} &= a_{1} b'_{1} + \dfrac{ b_{1} }{ a_{2} } = a_{1} b'_{1} - a_{1} \cdot \left( - \dfrac{ b_{1} }{ a_{1} a_{2} } \right) = a_{1} b'_{1} - a_{1} b'_{1} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{2} b'_{2} + b_{2} a'_{3} &= a_{2} b'_{2} + \dfrac{ b_{2} }{ a_{3} } = a_{2} b'_{2} - a_{2} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) = a_{2} b'_{2} - a_{2} b'_{2} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} c' + b_{1} b'_{2} + c a'_{3} &= a_{1} c' + b_{1} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) + \dfrac{ c }{ a_{3} } \\[10pt] &= a_{1} c' - \dfrac{ b_{1} b_{2} - a_{2} c }{ a_{2} a_{3} } \\[10pt] &= a_{1} c' - \dfrac{ 1 }{ a_{2} a_{3} } \left( b_{1} \left| \begin{array}{cc} 1 & c \\ 0 & b_{2} \\ \end{array} \right| + a_{2} \left| \begin{array}{cc} 0 & c \\ 1 & b_{2} \\ \end{array} \right| \right) \\[10pt] &= a_{1} c' - \dfrac{ 1 }{ a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c' - a_{1} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c' - a_{1} c' \\[10pt] &= 0 \, \boldsymbol{.} \end{aligned} \end{align}
   ${}$

5. $\textsf{[ 計算終了 ]}$

   ${}$
   よって，$A A' = E_{3}$ となるので，$A' = A^{-1}$ である。
   ${}$

$n = 4$ の場合

$\displaystyle A = \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline 0 & \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] $ と表す。このとき，$ \displaystyle A' = \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a'_{1} & \cellcolor{ lightgreen } b'_{1} & \cellcolor{ lightsalmon } c'_{1} & \cellcolor{ aquamarine } d' \\ \hline 0 & \cellcolor{ pink } a'_{2} & \cellcolor{ lightgreen } b'_{2} & \cellcolor{ lightsalmon } c'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} & \cellcolor{ lightgreen } b'_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a'_{4} \\ \end{array} \right] $ が $A$ の逆行列となるように成分を決めていく。
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1. $\textsf{[ 対角成分 $a'_{1} \, \boldsymbol{,} \ a'_{2} \, \boldsymbol{,} \ a'_{3} \, \boldsymbol{,} \ a'_{4}$ の決め方 ]}$

   ${}$
   $A'$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a'_{1} & b'_{1} & c'_{1} & d \\ \hline 0 & \cellcolor{ pink } a'_{2} & b'_{2} & c'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} & b'_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a'_{4} \\ \end{array} \right] $ を決めるには，$A$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} & d \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} & c_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & b_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] $ のそれぞれに着目し，
   \begin{align} \qquad \begin{aligned} a'_{1} = a_{1}^{-1} \, \boldsymbol{,} \qquad a'_{2} = a_{2}^{-1} \, \boldsymbol{,} \qquad a'_{3} = a_{3}^{-1} \, \boldsymbol{,} \qquad a'_{4} = a_{4}^{-1} \end{aligned} \end{align}
   と決める。
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2. $\textsf{[ 対角隣接成分 $b'_{1} \, \boldsymbol{,} \, b'_{2} \, \boldsymbol{,} \, b'_{3}$ の決め方 ]}$

   ${}$
   $A'$ の対角隣接成分 $ \displaystyle \left[ \begin{array}{c|c|c|c} a'_{1} & \cellcolor{ lightgreen } b'_{1} & c'_{1} & d \\ \hline 0 & a'_{2} & \cellcolor{ lightgreen } b'_{2} & c'_{2} \\ \hline 0 & 0 & a'_{3} & \cellcolor{ lightgreen } b'_{3} \\ \hline 0 & 0 & 0 & a'_{4} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(2-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & b'_{1} = (-1)^{1+2} \cdot \dfrac{ b_{1} }{ a_{1} a_{2} } = - \dfrac{ b_{1} }{ a_{1} a_{2} } \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(2-b)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & b'_{2} = (-1)^{2+3} \cdot \dfrac{ b_{2} }{ a_{2} a_{3} } = - \dfrac{ b_{2} }{ a_{2} a_{3} } \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(2-c)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & b'_{3} = (-1)^{3+4} \cdot \dfrac{ b_{3} }{ a_{3} a_{4} } = - \dfrac{ b_{3} }{ a_{3} a_{4} } \\[10pt] & \textsf{と決める。} \end{aligned} \end{align}
   ${}$

3. $\textsf{[ 成分 $c'_{2} \, \boldsymbol{,} \, c'_{2}$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 3 ) \, \boldsymbol{,} \ ( 2 \boldsymbol{,} \, 4 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c|c} a'_{1} & b'_{1} & \cellcolor{ lightsalmon } c'_{1} & d' \\ \hline 0 & a'_{2} & b'_{2} & \cellcolor{ lightsalmon } c'_{2} \\ \hline 0 & 0 & a'_{3} & b'_{3} \\ \hline 0 & 0 & 0 & a'_{4} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{1} = (-1)^{1+3} \cdot \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| = \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-b)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{1} = \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| \\[10pt] & \textsf{として分母を決める；} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-c)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{2} = (-1)^{2+4} \cdot \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| = \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-d)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{2} & b_{2} & c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & b_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{2} = \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{として分母を決める。} \end{aligned} \end{align}
   ${}$

4. $\textsf{[ $( 1 \boldsymbol{,} \, 4 )$ 成分 $d'$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 4 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c|c} a'_{1} & b'_{1} & c'_{1} & \cellcolor{ aquamarine } d' \\ \hline 0 & a'_{2} & b'_{2} & c'_{2} \\ \hline 0 & 0 & a'_{3} & b'_{3} \\ \hline 0 & 0 & 0 & a'_{4} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(4-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & d' = (-1)^{1+4} \cdot \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| = - \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(4-b)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} & d \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} & c_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & b_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & d' = - \dfrac{ 1 }{ \det A } \cdot \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| = - \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} } \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{として分母を決める。} \end{aligned} \end{align}
   ${}$

5. $\textsf{[ 成分計算 ]}$

   ${}$
   ゆえに，
   \begin{align} \qquad \begin{aligned} A A' &= \left[ \begin{array}{cccc} a_{1} & b_{1} & c_{1} & d \\ 0 & a_{2} & b_{2} & c_{2} \\ 0 & 0 & a_{3} & b_{3} \\ 0 & 0 & 0 & a_{4} \\ \end{array} \right] \left[ \begin{array}{cccc} a'_{1} & b'_{1} & c'_{1} & d' \\ 0 & a'_{2} & b'_{2} & c'_{2} \\ 0 & 0 & a'_{3} & b'_{3} \\ 0 & 0 & 0 & a'_{4} \\ \end{array} \right] \\[10pt] &= \left[ \begin{array}{c|c|c|c} a_{1} a'_{1} & a_{1} b'_{1} + b_{1} a'_{2} & a_{1} c'_{1} + b_{1} b'_{2} + c_{1} a'_{3} & a_{1} d' + b_{1} c'_{2} + c_{1} b'_{3} + d a'_{4} \\[5pt] \hline 0 & a_{2} a'_{2} & a_{2} b'_{2} + b_{2} a'_{3} & a_{2} c'_{2} + b_{2} b'_{3} + c_{2} a'_{4} \\[5pt] \hline 0 & 0 & a_{3} a'_{3} & a_{3} b'_{3} + b_{3} a'_{4} \\[5pt] \hline 0 & 0 & 0 & a_{4} a'_{4} \\ \end{array} \right] \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad & \begin{aligned} a_{1} a'_{1} = a_{2} a'_{2} = a_{3} a'_{3} = a_{4} a'_{4} = 1 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} b'_{1} + b_{1} a'_{2} &= a_{1} b'_{1} + \dfrac{ b_{1} }{ a_{2} } = a_{1} b'_{1} - a_{1} \cdot \left( - \dfrac{ b_{1} }{ a_{1} a_{2} } \right) = a_{1} b'_{1} - a_{1} b'_{1} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{2} b'_{2} + b_{2} a'_{3} &= a_{2} b'_{2} + \dfrac{ b_{2} }{ a_{3} } = a_{2} b'_{2} - a_{2} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) = a_{2} b'_{2} - a_{2} b'_{2} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{3} b'_{3} + b_{3} a'_{4} &= a_{3} b'_{3} + \dfrac{ b_{3} }{ a_{4} } = a_{3} b'_{3} - a_{3} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) = a_{3} b'_{3} - a_{3} b'_{3} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} c'_{1} + b_{1} b'_{2} + c_{1} a'_{3} &= a_{1} c'_{1} + b_{1} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) + \dfrac{ c_{1} }{ a_{3} } \\[10pt] &= a_{1} c'_{1} - \dfrac{ b_{1} b_{2} - a_{2} c_{1} }{ a_{2} a_{3} } \\[10pt] &= a_{1} c'_{1} - \dfrac{ 1 }{ a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c_{1} \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c'_{1} - a_{1} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c_{1} \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c'_{1} - a_{1} c'_{1} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[10pt] & \begin{aligned} a_{2} c'_{2} + b_{2} b'_{3} + c_{2} a'_{4} &= a_{2} c'_{2} + b_{2} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) + \dfrac{ c_{2} }{ a_{4} } \\[10pt] &= a_{2} c'_{2} - \dfrac{ b_{2} b_{3} - a_{3} c_{2} }{ a_{3} a_{4} } \\[10pt] &= a_{2} c'_{2} - \dfrac{ 1 }{ a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| \\[10pt] &= a_{2} c'_{2} - a_{2} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| \\[10pt] &= a_{2} c'_{2} - a_{2} c'_{2} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} d' + b_{1} c'_{2} + c_{1} b'_{3} + d a'_{4} &= a_{1} d' + b_{1} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| + c_{1} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) + \dfrac{ d }{ a_{4} } \\[10pt] &= a_{1} d' + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| - a_{2} b_{3} c_{1} + a_{2} a_{3} d \right) \\[10pt] &= a_{1} d' + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| - a_{2} \left| \begin{array}{cc} c_{1} & d \\ a_{3} & b_{3} \\ \end{array} \right| \right) \\[10pt] &= a_{1} d' + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{ccc} 1 & c_{1} & d \\ 0 & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| + a_{2} \left| \begin{array}{ccc} 0 & c_{1} & d \\ 1 & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| \right) \\[10pt] &= a_{1} d' - a_{1} \cdot \left( - \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} } \left| \begin{array}{ccc} b_{1} & c_{1} & d \\ a_{2} & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| \right) \\[10pt] &= a_{1} d' - a_{1} d' \\[10pt] &= 0 \, \boldsymbol{.} \end{aligned} \end{align}
   ${}$

6. $\textsf{[ 計算終了 ]}$

   ${}$
   よって，$A A' = E_{4}$ となるので，$A' = A^{-1}$ である。
   ${}$

$n = 5$ の場合

$\displaystyle A = \left[ \begin{array}{c|c|c|c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} & \cellcolor{ gold } e \\ \hline 0 & \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \hline 0 & 0 & 0 & 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] $ と表す。このとき，$ \displaystyle A' = \left[ \begin{array}{c|c|c|c|c} \cellcolor{ pink } a'_{1} & \cellcolor{ lightgreen } b'_{1} & \cellcolor{ lightsalmon } c'_{1} & \cellcolor{ aquamarine } d'_{1} & \cellcolor{ gold } e' \\ \hline 0 & \cellcolor{ pink } a'_{2} & \cellcolor{ lightgreen } b'_{2} & \cellcolor{ lightsalmon } c'_{2} & \cellcolor{ aquamarine } d'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} & \cellcolor{ lightgreen } b'_{3} & \cellcolor{ lightsalmon } c'_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a'_{4} & \cellcolor{ lightgreen } b'_{4} \\ \hline 0 & 0 & 0 & 0 & \cellcolor{ pink } a'_{5} \\ \end{array} \right] $ が $A$ の逆行列となるように成分を決めていく。
${}$

1. $\textsf{[ 対角成分 $a'_{1} \, \boldsymbol{,} \ a'_{2} \, \boldsymbol{,} \ a'_{3} \, \boldsymbol{,} \ a'_{4} \, \boldsymbol{,} \ a'_{5}$ の決め方 ]}$

   ${}$
   $A'$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} \cellcolor{ pink } a'_{1} & b'_{1} & c'_{1} & d'_{1} & e' \\ \hline 0 & \cellcolor{ pink } a'_{2} & b'_{2} & c'_{2} & d'_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a'_{3} & b'_{3} & c'_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a'_{4} & b'_{4} \\ \hline 0 & 0 & 0 & 0 & \cellcolor{ pink } a'_{5} \\ \end{array} \right] $ を決めるには，$A$ の各対角成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} & d_{1} & e \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} & c_{2} & d_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & b_{3} & c_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} & b_{4} \\ \hline 0 & 0 & 0 & 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] $ のそれぞれに着目し，
   \begin{align} \qquad \begin{aligned} a'_{1} = a_{1}^{-1} \, \boldsymbol{,} \qquad a'_{2} = a_{2}^{-1} \, \boldsymbol{,} \qquad a'_{3} = a_{3}^{-1} \, \boldsymbol{,} \qquad a'_{4} = a_{4}^{-1} \, \boldsymbol{,} \qquad a'_{5} = a_{5}^{-1} \end{aligned} \end{align}
   と決める。
   ${}$

2. $\textsf{[ 対角隣接成分 $b'_{1} \, \boldsymbol{,} \, b'_{2} \, \boldsymbol{,} \, b'_{3} \, \boldsymbol{,} \, b'_{4}$ の決め方 ]}$

   ${}$
   $A'$ の対角隣接成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} a'_{1} & \cellcolor{ lightgreen } b'_{1} & c'_{1} & d'_{1} & e' \\ \hline 0 & a'_{2} & \cellcolor{ lightgreen } b'_{2} & c'_{2} & d'_{2} \\ \hline 0 & 0 & a'_{3} & \cellcolor{ lightgreen } b'_{3} & c'_{3} \\ \hline 0 & 0 & 0 & a'_{4} & \cellcolor{ lightgreen } b'_{4} \\ \hline 0 & 0 & 0 & 0 & a'_{5} \\ \end{array} \right] $ を決めるには，
   ${}$
   $A$ のブロック行列 $ \displaystyle \left[ \begin{array}{c|c} \cellcolor{ pink } a_{1} & \cellcolor{ lightgreen } b_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} \\ \end{array} \right] \, \boldsymbol{,} \ \ \left[ \begin{array}{c|c} \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \, \boldsymbol{,} \ \ \left[ \begin{array}{c|c} \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \, \boldsymbol{,} \ \ \left[ \begin{array}{c|c} \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \hline 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] $ のそれぞれに着目し，
   ${}$
   \begin{align} \qquad & \begin{aligned} & b'_{1} = (-1)^{1+2} \cdot \dfrac{ b_{1} }{ a_{1} a_{2} } = - \dfrac{ b_{1} }{ a_{1} a_{2} } \, \boldsymbol{,} \\[10pt] & b'_{2} = (-1)^{2+3} \cdot \dfrac{ b_{2} }{ a_{2} a_{3} } = - \dfrac{ b_{2} }{ a_{2} a_{3} } \, \boldsymbol{,} \\[10pt] & b'_{3} = (-1)^{3+4} \cdot \dfrac{ b_{3} }{ a_{3} a_{4} } = - \dfrac{ b_{3} }{ a_{3} a_{4} } \, \boldsymbol{,} \\[10pt] & b'_{4} = (-1)^{4+5} \cdot \dfrac{ b_{4} }{ a_{4} a_{5} } = - \dfrac{ b_{4} }{ a_{4} a_{5} } \end{aligned} \end{align}
   と決める。
   ${}$

3. $\textsf{[ 成分 $c'_{2} \, \boldsymbol{,} \, c'_{2} \, \boldsymbol{,} \, c'_{3}$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 3 ) \, \boldsymbol{,} \ ( 2 \boldsymbol{,} \, 4 ) \, \boldsymbol{,} \ ( 3 \boldsymbol{,} \, 5 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} a'_{1} & b'_{1} & \cellcolor{ lightsalmon } c'_{1} & d'_{1} & e' \\ \hline 0 & a'_{2} & b'_{2} & \cellcolor{ lightsalmon } c'_{2} & d'_{2} \\ \hline 0 & 0 & a'_{3} & b'_{3} & \cellcolor{ lightsalmon } c'_{3} \\ \hline 0 & 0 & 0 & a'_{4} & b'_{4} \\ \hline 0 & 0 & 0 & 0 & a'_{5} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right] \quad \textsf{および} \quad \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{1} = (-1)^{1+3} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| = \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} \\ \end{array} \right| \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-b)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right] \quad \textsf{および} \quad \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{2} & b_{2} & c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & b_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{2} = (-1)^{2+4} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| = \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(3-c)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c} \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right] \quad \textsf{および} \quad \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c} \cellcolor{ pink } a_{3} & b_{3} & c_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} & b_{4} \\ \hline 0 & 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] \quad \textsf{に着目し，} \quad \\[10pt] & c'_{3} = (-1)^{3+5} \cdot \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \cdot \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| = \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left| \begin{array}{c|c} \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| \\[10pt] & \textsf{と決める。} \end{aligned} \end{align}
   ${}$

4. $\textsf{[ 成分 $d'_{1} \, \boldsymbol{,} \, d'_{2}$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 4 ) \, \boldsymbol{,} \, ( 2 \boldsymbol{,} \, 5 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} a'_{1} & b'_{1} & c'_{1} & \cellcolor{ aquamarine } d'_{1} & e' \\ \hline 0 & a'_{2} & b'_{2} & c'_{2} & \cellcolor{ aquamarine } d'_{2} \\ \hline 0 & 0 & a'_{3} & b'_{3} & c'_{3} \\ \hline 0 & 0 & 0 & a'_{4} & b'_{4} \\ \hline 0 & 0 & 0 & 0 & a'_{5} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(4-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right] \quad \textsf{および} \quad \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} & d_{1} \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} & c_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & b_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & d'_{1} = (-1)^{1+4} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} } \cdot \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| = - \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} } \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} \\ \end{array} \right| \\[10pt] & \textsf{と決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(4-b)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right] \quad \textsf{および} \quad \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c|c} \cellcolor{ pink } a_{2} & b_{2} & c_{2} & d_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & b_{3} & c_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} & b_{4} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & d'_{2} = (-1)^{2+5} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \cdot \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| = - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{c|c|c} \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| \\[10pt] & \textsf{と決める。} \end{aligned} \end{align}
   ${}$

5. $\textsf{[ $( 1 \boldsymbol{,} \, 5 )$ 成分 $e'$ の決め方 ]}$

   ${}$
   $A'$ の $( 1 \boldsymbol{,} \, 5 )$ 成分 $ \displaystyle \left[ \begin{array}{c|c|c|c|c} a'_{1} & b'_{1} & c'_{1} & d'_{1} & \cellcolor{ gold } e' \\ \hline 0 & a'_{2} & b'_{2} & c'_{2} & d'_{2} \\ \hline 0 & 0 & a'_{3} & b'_{3} & c'_{3} \\ \hline 0 & 0 & 0 & a'_{4} & b'_{4} \\ \hline 0 & 0 & 0 & 0 & a'_{5} \\ \end{array} \right] $ を決めるには，
   ${}$
   \begin{align} \begin{aligned} \textsf{(5-a)} \quad & \textsf{$A$ のブロック行列} \quad \left[ \begin{array}{c|c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} & \cellcolor{ gold } e \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & e' = (-1)^{1+5} \cdot \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} & \cellcolor{ gold } e \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| = \dfrac{ 1 }{ \fbox{$\phantom{\det A}$} } \cdot \left| \begin{array}{c|c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} & \cellcolor{ gold } e \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| \\[10pt] & \textsf{として分母以外を決め，} \end{aligned} \end{align}
   ${}$
   \begin{align} \begin{aligned} \textsf{(5-b)} \quad & \textsf{$A$ の対角成分} \quad \left[ \begin{array}{c|c|c|c|c} \cellcolor{ pink } a_{1} & b_{1} & c_{1} & d_{1} & e \\ \hline 0 & \cellcolor{ pink } a_{2} & b_{2} & c_{2} & d_{2} \\ \hline 0 & 0 & \cellcolor{ pink } a_{3} & b_{3} & c_{3} \\ \hline 0 & 0 & 0 & \cellcolor{ pink } a_{4} & b_{4} \\ \hline 0 & 0 & 0 & 0 & \cellcolor{ pink } a_{5} \\ \end{array} \right] \quad \textsf{に着目し，} \\[10pt] & e' = \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{c|c|c|c} \cellcolor{ lightgreen } b_{1} & \cellcolor{ lightsalmon } c_{1} & \cellcolor{ aquamarine } d_{1} & \cellcolor{ gold } e \\ \hline \cellcolor{ pink } a_{2} & \cellcolor{ lightgreen } b_{2} & \cellcolor{ lightsalmon } c_{2} & \cellcolor{ aquamarine } d_{2} \\ \hline 0 & \cellcolor{ pink } a_{3} & \cellcolor{ lightgreen } b_{3} & \cellcolor{ lightsalmon } c_{3} \\ \hline 0 & 0 & \cellcolor{ pink } a_{4} & \cellcolor{ lightgreen } b_{4} \\ \end{array} \right| \\[10pt] & \textsf{として分母を決める。} \end{aligned} \end{align}
   ${}$

6. $\textsf{[ 成分計算 ]}$

   ${}$
   ゆえに，
   \begin{align} \qquad \begin{aligned} A A' &= \left[ \begin{array}{ccccc} a_{1} & b_{1} & c_{1} & d_{1} & e \\ 0 & a_{2} & b_{2} & c_{2} & d_{2} \\ 0 & 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & 0 & a_{4} & b_{4} \\ 0 & 0 & 0 & 0 & a_{5} \end{array} \right] \left[ \begin{array}{cccc} a'_{1} & b'_{1} & c'_{1} & d'_{1} & e' \\ 0 & a'_{2} & b'_{2} & c'_{2} & d'_{2} \\ 0 & 0 & a'_{3} & b'_{3} & c'_{3} \\ 0 & 0 & 0 & a'_{4} & b'_{4} \\ 0 & 0 & 0 & 0 & a'_{5} \end{array} \right] \\[10pt] &= \left[ \begin{array}{c|c|c|c|c} a_{1} a'_{1} & a_{1} b'_{1} + b_{1} a'_{2} & a_{1} c'_{1} + b_{1} b'_{2} + c_{1} a'_{3} & a_{1} d'_{1} + b_{1} c'_{2} + c_{1} b'_{3} + d_{1} a'_{4} & a_{1} e' + b_{1} d'_{2} + c_{1} c'_{3} + d_{1} b'_{4} + e a'_{5} \\[5pt] \hline 0 & a_{2} a'_{2} & a_{2} b'_{2} + b_{2} a'_{3} & a_{2} c'_{2} + b_{2} b'_{3} + c_{2} a'_{4} & a_{2} d'_{2} + b_{2} c'_{3} + c_{2} b'_{4} + d_{2} a'_{5} \\[5pt] \hline 0 & 0 & a_{3} a'_{3} & a_{3} b'_{3} + b_{3} a'_{4} & a_{3} c'_{3} + b_{3} b'_{4} + c_{3} a'_{5} \\[5pt] \hline 0 & 0 & 0 & a_{4} a'_{4} & a_{4} b'_{4} + b_{4} a'_{5} \\[5pt] \hline 0 & 0 & 0 & 0 & a_{5} a'_{5} \\ \end{array} \right] \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad & \begin{aligned} a_{1} a'_{1} = a_{2} a'_{2} = a_{3} a'_{3} = a_{4} a'_{4} = a_{5} a'_{5} = 1 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} b'_{1} + b_{1} a'_{2} &= a_{1} b'_{1} + \dfrac{ b_{1} }{ a_{2} } = a_{1} b'_{1} - a_{1} \cdot \left( - \dfrac{ b_{1} }{ a_{1} a_{2} } \right) = a_{1} b'_{1} - a_{1} b'_{1} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{2} b'_{2} + b_{2} a'_{3} &= a_{2} b'_{2} + \dfrac{ b_{2} }{ a_{3} } = a_{2} b'_{2} - a_{2} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) = a_{2} b'_{2} - a_{2} b'_{2} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{3} b'_{3} + b_{3} a'_{4} &= a_{3} b'_{3} + \dfrac{ b_{3} }{ a_{4} } = a_{3} b'_{3} - a_{3} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) = a_{3} b'_{3} - a_{3} b'_{3} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{4} b'_{4} + b_{4} a'_{5} &= a_{4} b'_{4} + \dfrac{ b_{4} }{ a_{5} } = a_{4} b'_{4} - a_{4} \cdot \left( - \dfrac{ b_{4} }{ a_{4} a_{5} } \right) = a_{4} b'_{4} - a_{4} b'_{4} = 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} c'_{1} + b_{1} b'_{2} + c_{1} a'_{3} &= a_{1} c'_{1} + b_{1} \cdot \left( - \dfrac{ b_{2} }{ a_{2} a_{3} } \right) + \dfrac{ c_{1} }{ a_{3} } \\[10pt] &= a_{1} c'_{1} - \dfrac{ b_{1} b_{2} - a_{2} c_{1} }{ a_{2} a_{3} } \\[10pt] &= a_{1} c'_{1} - \dfrac{ 1 }{ a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c_{1} \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c'_{1} - a_{1} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} } \left| \begin{array}{cc} b_{1} & c_{1} \\ a_{2} & b_{2} \\ \end{array} \right| \\[10pt] &= a_{1} c'_{1} - a_{1} c'_{1} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{2} c'_{2} + b_{2} b'_{3} + c_{2} a'_{4} &= a_{2} c'_{2} + b_{2} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) + \dfrac{ c_{2} }{ a_{4} } \\[10pt] &= a_{2} c'_{2} - \dfrac{ b_{2} b_{3} - a_{3} c_{2} }{ a_{3} a_{4} } \\[10pt] &= a_{2} c'_{2} - \dfrac{ 1 }{ a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| \\[10pt] &= a_{2} c'_{2} - a_{2} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| \\[10pt] &= a_{2} c'_{2} - a_{2} c'_{2} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{3} c'_{3} + b_{3} b'_{4} + c_{3} a'_{5} &= a_{3} c'_{3} + b_{3} \cdot \left( - \dfrac{ b_{4} }{ a_{4} a_{5} } \right) + \dfrac{ c_{3} }{ a_{5} } \\[10pt] &= a_{3} c'_{3} - \dfrac{ b_{3} b_{4} - a_{4} c_{3} }{ a_{4} a_{5} } \\[10pt] &= a_{3} c'_{3} - \dfrac{ 1 }{ a_{4} a_{5} } \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| \\[10pt] &= a_{3} c'_{3} - a_{3} \cdot \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| \\[10pt] &= a_{3} c'_{3} - a_{3} c'_{3} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{1} d'_{1} + b_{1} c'_{2} + c_{1} b'_{3} + d_{1} a'_{4} &= a_{1} d'_{1} + b_{1} \cdot \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| + c_{1} \cdot \left( - \dfrac{ b_{3} }{ a_{3} a_{4} } \right) + \dfrac{ d_{1} }{ a_{4} } \\[10pt] &= a_{1} d'_{1} + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| - a_{2} b_{3} c_{1} + a_{2} a_{3} d_{1} \right) \\[10pt] &= a_{1} d'_{1} + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{cc} b_{2} & c_{2} \\ a_{3} & b_{3} \\ \end{array} \right| - a_{2} \left| \begin{array}{cc} c_{1} & d_{1} \\ a_{3} & b_{3} \\ \end{array} \right| \right) \\[10pt] &= a_{1} d'_{1} + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left( b_{1} \left| \begin{array}{ccc} 1 & c_{1} & d_{1} \\ 0 & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| + a_{2} \left| \begin{array}{ccc} 0 & c_{1} & d_{1} \\ 1 & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| \right) \\[10pt] &= a_{1} d'_{1} + \dfrac{ 1 }{ a_{2} a_{3} a_{4} } \left| \begin{array}{ccc} b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| \\[10pt] &= a_{1} d'_{1} - a_{1} \cdot \left( - \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} } \left| \begin{array}{ccc} b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} \\ 0 & a_{3} & b_{3} \end{array} \right| \right) \\[10pt] &= a_{1} d'_{1} - a_{1} d'_{1} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[15pt] & \begin{aligned} a_{2} d'_{2} + b_{2} c'_{3} + c_{2} b'_{4} + d_{2} a'_{5} &= a_{2} d'_{2} + b_{2} \cdot \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| + c_{2} \cdot \left( - \dfrac{ b_{4} }{ a_{4} a_{5} } \right) + \dfrac{ d_{2} }{ a_{5} } \\[10pt] &= a_{2} d'_{2} + \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left( b_{2} \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| - a_{3} b_{4} c_{2} + a_{3} a_{4} d_{2} \right) \\[10pt] &= a_{2} d'_{2} + \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left( b_{2} \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| - a_{3} \left| \begin{array}{cc} c_{2} & d_{2} \\ a_{4} & b_{4} \\ \end{array} \right| \right) \\[10pt] &= a_{2} d'_{2} + \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left( b_{2} \left| \begin{array}{ccc} 1 & c_{2} & d_{2} \\ 0 & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| + a_{3} \left| \begin{array}{ccc} 0 & c_{2} & d_{2} \\ 1 & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| \right) \\[10pt] &= a_{2} d'_{2} + \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left| \begin{array}{ccc} b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| \\[10pt] &= a_{2} d'_{2} - a_{2} \cdot \left( - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{ccc} b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| \right) \\[10pt] &= a_{2} d'_{2} - a_{2} d'_{2} \\[10pt] &= 0 \, \boldsymbol{,} \end{aligned} \\[20pt] & \begin{aligned} a_{1} e' + b_{1} d'_{2} + c_{1} c'_{3} + d_{1} b'_{4} + e a'_{5} &= a_{1} e' + b_{1} \cdot \left( - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{ccc} b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| \right) + c_{1} \cdot \dfrac{ 1 }{ a_{3} a_{4} a_{5} } \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| + d_{1} \cdot \left( - \dfrac{ b_{4} }{ a_{4} a_{5} } \right) + \dfrac{ e }{ a_{5} } \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( b_{1} \left| \begin{array}{ccc} b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \\ \end{array} \right| - c_{1} a_{2} \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| + d_{1} a_{2} a_{3} b_{4} - e a_{2} a_{3} a_{4} \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( b_{1} \left| \begin{array}{cccc} 1 & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| - c_{1} a_{2} \left| \begin{array}{cc} b_{3} & c_{3} \\ a_{4} & b_{4} \\ \end{array} \right| + a_{2} a_{3} \left| \begin{array}{cc} d_{1} & e \\ a_{4} & b_{4} \\ \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| - c_{1} a_{2} \left| \begin{array}{ccc} 1 & d_{1} & e \\ 0 & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \end{array} \right| - a_{2} a_{3} \left| \begin{array}{ccc} 0 & d_{1} & e \\ 1 & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| - a_{2} \left| \begin{array}{ccc} c_{1} & d_{1} & e \\ 0 & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \end{array} \right| - a_{2} \left| \begin{array}{ccc} 0 & d_{1} & e \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| - a_{2} \left| \begin{array}{ccc} c_{1} & d_{1} & e \\ a_{3} & b_{3} & c_{3} \\ 0 & a_{4} & b_{4} \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| + a_{2} \left| \begin{array}{cccc} 0 & c_{1} & d_{1} & e \\ 1 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left( \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ 0 & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| + \left| \begin{array}{cccc} 0 & c_{1} & d_{1} & e \\ a_{2} & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| \right) \\[10pt] &= a_{1} e' - \dfrac{ 1 }{ a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ a_{2} & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| \\[10pt] &= a_{1} e' - a_{1} \cdot \dfrac{ 1 }{ a_{1} a_{2} a_{3} a_{4} a_{5} } \left| \begin{array}{cccc} b_{1} & c_{1} & d_{1} & e \\ a_{2} & b_{2} & c_{2} & d_{2} \\ 0 & a_{3} & b_{3} & c_{3} \\ 0 & 0 & a_{4} & b_{4} \end{array} \right| \\[10pt] &= a_{1} e' - a_{1} e' \\[10pt] &= 0 \, \boldsymbol{.} \end{aligned} \end{align}
   ${}$

7. $\textsf{[ 計算終了 ]}$

   ${}$
   よって，$A A' = E_{5}$ となるので，$A' = A^{-1}$ である。
   ${}$

$n \geq 4$ の場合

$ \displaystyle \begin{aligned} A_{n} := A = \left[ \begin{array}{ccc} a_{1 \, , \, 1} & \cdots & a_{1 \, , \, n} \\ {} & \ddots & \vdots \\ {\Huge\boldsymbol{0}} & & a_{n \, , \, n} \\ \end{array} \right] \end{aligned} $ と表す。

また，各 $i \in \left\{ 1 \boldsymbol{,} \, \boldsymbol{\dots} \boldsymbol{,} \, n \right\}$ および 各 $j \in \left\{ 0 \boldsymbol{,} \, \boldsymbol{\dots} \boldsymbol{,} \, n-1 \right\}$ に対して，$a'_{i \, , \, i+j} \in \mathbb{C}$ をそれぞれ，
${}$
\begin{align} \quad & \begin{aligned} & \boldsymbol{\textsf{[\,i\,]}} \quad j = 0 \quad \textsf{ならば，} \\[10pt] &{\qquad \quad} a'_{ i \, , \, i } := \dfrac{1}{ a_{ i \, , \, i } } \ \boldsymbol{;} \end{aligned} \\[20pt] & \begin{aligned} & \boldsymbol{\textsf{[\,ii\,]}} \quad j = 1 \quad \textsf{かつ} \quad 1 \leq i \leq n - 1 \quad \textsf{ならば，} \\[10pt] &{\qquad \quad} a'_{ i \, , \, i+1 } := - \dfrac{ a_{ i \, , \, i+1 } }{ a_{ i \, , \, i } \cdot a_{ i+1 \, , \, i+1 } } \ \boldsymbol{;} \end{aligned} \\[20pt] & \begin{aligned} & \boldsymbol{\textsf{[\,iii\,]}} \quad j = 2 \quad \textsf{かつ} \quad 1 \leq i \leq n - 2 \quad \textsf{ならば，} \\[10pt] &{\qquad \quad} a'_{ i \, , \, i+2 } := \dfrac{ 1 }{ \prod\limits_{\nu=0}^{2} a_{ i+\nu \, , \, i+\nu } } \left| \begin{array}{cc} a_{ i \, ,\, i+1 } & a_{ i \, , \, i+2 } \\ a_{ i+1 \, , \, i+1 } & a_{ i+1 \, , \, i+2 } \\ \end{array} \right| \ \boldsymbol{;} \end{aligned} \\[20pt] & \begin{aligned} & \boldsymbol{\textsf{[\,iv\,]}} \quad 3 \leq j \leq n - 1 \quad \textsf{かつ} \quad 1 \leq i \leq n - j \quad \textsf{ならば，} \\[10pt] &{\qquad \quad} a'_{ i \, , \, i+j } := \dfrac{ \left( -1 \right)^{j} }{ \prod\limits_{\nu=0}^{j} a_{ i+\nu \, , \, i+\nu } } \left| \begin{array}{cccccc} a_{ i \, , \, i+1 } & a_{ i \, , \, i+2 } & \cdots & a_{ i \, , \, i+j-2 } & a_{ i \, , \, i+j-1 } & a_{ i \, , \, i+j } \\ a_{ i+1 \, , \, i+1 } & a_{ i+1 \, , \, i+2 } & \cdots & a_{ i+1 \, , \, i+j-2 } & a_{ i+1 \, , \, i+j-1 } & a_{ i+1 \, , \, i+j } \\ 0 & a_{ i+2 \, , \, i+2 } & \cdots & a_{ i+2 \, , \, i+j-2 } & a_{ i+2 \, , \, i+j-1 } & a_{ i+2 \, , \, i+j } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ i+j-2 \, , \, i+j-2 } & a_{ i+j-2 \, , \, i+j-1 } & a_{ i+j-2 \, , \, i+j } \\ {\Huge\boldsymbol{0}} & & & 0 & a_{ i+j-1 \, , \, i+j-1 } & a_{ i+j-1 \, , \, i+j } \\ \end{array} \right| \end{aligned} \end{align}
として定め，
\begin{align} \qquad \begin{aligned} A'_{n} := \left[ \begin{array}{ccc} a'_{1 \, , \, 1} & \cdots & a'_{1 \, , \, n} \\ {} & \ddots & \vdots \\ {\Huge\boldsymbol{0}} & {} & a'_{n \, , \, n} \\ \end{array} \right] \end{aligned} \end{align}
とする。

---

帰納的に，$A'_{n} = A_{n}^{-1}$ となることを示していく。

$N \in \mathbb{Z} \cap \left[ 4 \boldsymbol{,} \, \infty \right)$ が与えられたとする。

$n = N$ の場合に $A_{n} A'_{n} = E_{n}$ が成立すると仮定する。このとき，

\begin{align} \qquad \begin{aligned} A_{N+1} A'_{N+1} &= \left[ \begin{array}{c|c} {} & a_{1 , \, N+1} \\ \ \ {\Huge A_{N}} \ \ & \vdots \\ {} & a_{N , \, N+1} \\[5pt] \hline \boldsymbol{0}^{\mathsf{T}} & a_{N+1 , \, N+1} \\ \end{array} \right] \left[ \begin{array}{c|c} {} & a'_{1 , \, N+1} \\ \ \ {\Huge A'_{N}} \ \ & \vdots \\ {} & a'_{N , \, N+1} \\[5pt] \hline \boldsymbol{0}^{\mathsf{T}} & a'_{N+1 , \, N+1} \\ \end{array} \right] \\[15pt] &= \left[ \begin{array}{ccc|c} a_{1 \, , \, 1} & \cdots & a_{1 , \, N} & a_{1 , \, N+1} \\ {} & \ddots & \vdots & \vdots \\ {\Huge \boldsymbol{0}} & {} & a_{N , \, N} & a_{N , \, N+1} \\[5pt] \hline 0 & \cdots & 0 & a_{N+1 , \, N+1} \\ \end{array} \right] \left[ \begin{array}{ccc|c} a'_{1 \, , \, 1} & \cdots & a'_{1 , \, N} & a'_{1 , \, N+1} \\ {} & \ddots & \vdots & \vdots \\ {\Huge \boldsymbol{0}} & {} & a'_{N , \, N} & a'_{N , \, N+1} \\[5pt] \hline 0 & \cdots & 0 & a'_{N+1 , \, N+1} \\ \end{array} \right] \\[15pt] &= \left[ \begin{array}{c|c} {} & \sum\limits_{\mu = 1}^{N+1} a_{ 1 , \, \mu } \ a'_{ \mu , \, N+1 } \\ \ \ {\Huge A_{N} A'_{N}} \ \ & \vdots \\ {} & \sum\limits_{\mu = N}^{N+1} a_{ N , \, \mu } \ a'_{ \mu , \, N+1 } \\[5pt] \hline \boldsymbol{0}^{\mathsf{T}} & a_{N+1 , \, N+1} \ a'_{N+1 , \, N+1} \\ \end{array} \right] \\[15pt] &= \left[ \begin{array}{c|c} {} & \sum\limits_{\mu = 1}^{N+1} a_{ 1 , \, \mu } \ a'_{ \mu , \, N+1 } \\ \ \ {\Huge E_{N}} \ \ & \vdots \\ {} & \sum\limits_{\mu = N}^{N+1} a_{ N , \, \mu } \ a'_{ \mu , \, N+1 } \\[5pt] \hline \boldsymbol{0}^{\mathsf{T}} & 1 \\ \end{array} \right] \ \boldsymbol{.} \end{aligned} \end{align}
${}$
また，各 $k \in \left\{ \, 1 \boldsymbol{,} \, \boldsymbol{\dots} \boldsymbol{,} \, N \, \right\}$ が与えられたとする。このとき，
${}$

1. $k = N$ の場合：
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= \sum_{\mu = N}^{N+1} a_{ N , \, \mu } \ a'_{ \mu , \, N+1 } \\[10pt] &= a_{ N , \, N } \ a'_{ N , \, N+1 } + a_{ N , \, N+1 } \ a'_{ N+1 , \, N+1 } \\[10pt] &= a_{ N , \, N } \ a'_{ N , \, N+1 } + \dfrac{ a_{ N , \, N+1 } }{ a_{ N+1 , \, N+1 } } \\[10pt] &= a_{ N , \, N } \ a'_{ N , \, N+1 } - a_{ N , \, N } \cdot \left( - \dfrac{ a_{ N , \, N+1 } }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \right) \\[10pt] &= a_{ N , \, N } \ a'_{ N , \, N+1 } - a_{ N , \, N } \ a'_{ N , \, N+1 } \\[10pt] &= 0 \ \boldsymbol{;} \end{aligned} \end{align}
   ${}$
2. $k = N - 1$ の場合：
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= \sum_{\mu = N-1}^{N+1} a_{ N-1 , \, \mu } \ a'_{ \mu , \, N+1 } \\[5pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } + a_{ N-1 , \, N } \ a'_{ N , \, N+1 } + a_{ N-1 , \, N+1 } \ a'_{ N+1 , \, N+1 } \\[10pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } + a_{ N-1 , \, N } \cdot \left( - \dfrac{ a_{ N , \, N+1 } }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \right) + \dfrac{ a_{ N-1 , \, N+1 } }{ a_{ N+1 , \, N+1 } } \\[10pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } - \dfrac{ 1 }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left( a_{ N-1 , \, N } \cdot a_{ N , \, N+1 } - a_{ N , \, N } \cdot a_{ N-1 , \, N+1 } \right) \\[10pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } - \dfrac{ 1 }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left| \begin{array}{cc} a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } - a_{ N-1 , \, N-1 } \cdot \dfrac{ 1 }{ a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left| \begin{array}{cc} a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } - a_{ N-1 , \, N-1 } \ a'_{ N-1 , \, N+1 } \\[10pt] &= 0 \ \boldsymbol{;} \end{aligned} \end{align}
   ${}$
3. $k = N - 2$ の場合：
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= \sum_{\mu = N-2}^{N+1} a_{ N-2 , \, \mu } \ a'_{ \mu , \, N+1 } \\[5pt] &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } + a_{ N-2 , \, N-1 } \ a'_{ N-1 , \, N+1 } + a_{ N-2 , \, N } \ a'_{ N , \, N+1 } + a_{ N-2 , \, N+1 } \ a'_{ N+1 , \, N+1 } \\[10pt] &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } \\[10pt] &{\qquad \qquad} + \dfrac{ a_{ N-2 , \, N-1 } }{ a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left| \begin{array}{cc} a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - \dfrac{ a_{ N-2 , \, N } \cdot a_{ N , \, N+1 } }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } + \dfrac{ a_{ N-2 , \, N+1 } }{ a_{ N+1 , \, N+1 } } \\[10pt] &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } \\[10pt] &{\qquad \qquad} + \dfrac{ 1 }{ a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left( \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - a_{ N-1 , \, N-1 } \cdot a_{ N-2 , \, N } \cdot a_{ N , \, N+1 } + a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N-2 , \, N+1 } \right) \boldsymbol{,} \end{aligned} \end{align}
   ${}$
   \begin{align} \qquad \begin{aligned} & \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - a_{ N-1 , \, N-1 } \cdot a_{ N-2 , \, N } \cdot a_{ N , \, N+1 } + a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N-2 , \, N+1 } \\[10pt] &{\qquad \qquad}= \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - a_{ N-1 \, , \, N-1 } \cdot \left| \begin{array}{ccc} a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &{\qquad \qquad}= \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| + \left| \begin{array}{ccc} 0 & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &{\qquad \qquad}= \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1, \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } + \dfrac{ 1 }{ a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1, \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } - a_{ N-2 , \, N-2 } \cdot \left( - \dfrac{ 1 }{ a_{ N-2 , \, N-2 } \cdot a_{ N-1 , \, N-1 } \cdot a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1, \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \right) \\[10pt] &= a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } - a_{ N-2 , \, N-2 } \ a'_{ N-2 , \, N+1 } \\[10pt] &= 0 \ \boldsymbol{;} \end{aligned} \end{align}
   ${}$
4. $k \leq N - 3$ の場合：
   ${}$
   数学的帰納法を用いる。
   まず，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ k , \, N } \ a'_{ N , \, N+1 } + a_{ k , \, N+1 } \ a'_{ N+1 , \, N+1 } \\[10pt] &= a_{ k , \, N } \cdot \left( - \dfrac{ a_{ N , \, N+1 } }{ a_{ N , \, N } \cdot a_{ N+1 , \, N+1 } } \right) + \dfrac{a_{ k , \, N+1 }}{a_{ N+1 , \, N+1 }} \\[10pt] &= \dfrac{ -1 }{ \prod\limits_{\nu = 0}^{ 1 } a_{ N+\nu , \, N+\nu } } \left| \begin{array}{cc} a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \ \boldsymbol{,} \end{aligned} \end{align}
   ${}$
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N-1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ k , \, N-1 } \ a'_{ N-1 , \, N+1 } + \sum_{\mu = N}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } \\[10pt] &= a_{ k , \, N-1 } \cdot \dfrac{ \left( -1 \right)^{2} }{ \prod\limits_{\nu = 0}^{ 2 } a_{ N-1+\nu , \, N-1+\nu } } \left| \begin{array}{cc} a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| + \dfrac{ -1 }{ \prod\limits_{\nu = 0}^{ 1 } a_{ N+\nu , \, N+\nu } } \left| \begin{array}{cc} a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= \dfrac{ 1 }{ \prod\limits_{\nu = 0}^{ 2 } a_{ N-1+\nu , \, N-1+\nu } } \left( a_{ k , \, N-1 } \cdot \left| \begin{array}{cc} a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - a_{ N-1 , \, N-1 } \cdot \left| \begin{array}{cc} a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \right) \\[10pt] &= \dfrac{ 1 }{ \prod\limits_{\nu = 0}^{ 2 } a_{ N-1+\nu , \, N-1+\nu } } \left( \left| \begin{array}{ccc} a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ 0 & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| + \left| \begin{array}{ccc} 0 & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \right) \\[10pt] &= \dfrac{ 1 }{ \prod\limits_{\nu = 0}^{ 2 } a_{ N-1+\nu , \, N-1+\nu } } \left| \begin{array}{ccc} a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \end{aligned} \end{align}
   となることから，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N-2}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ k , \, N-2 } \ a'_{ N-2 , \, N+1 } + \sum_{\mu = N-1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } \\[10pt] &= a_{ k , \, N-2 } \cdot \dfrac{ \left( -1 \right)^{3} }{ \prod\limits_{\nu = 0}^{ 3 } a_{ N-2+\nu , \, N-2+\nu } } \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| + \dfrac{ 1 }{ \prod\limits_{\nu = 0}^{ 2 } a_{ N-1+\nu , \, N-1+\nu } } \left| \begin{array}{ccc} a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= \dfrac{ -1 }{ \prod\limits_{\nu = 0}^{ 3 } a_{ N-2+\nu , \, N-2+\nu } } \left( a_{ k , \, N-2 } \cdot \left| \begin{array}{ccc} a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| - a_{ N-2 , \, N-2 } \cdot \left| \begin{array}{ccc} a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \right) \\[10pt] &= \dfrac{ -1 }{ \prod\limits_{\nu = 0}^{ 3 } a_{ N-2+\nu , \, N-2+\nu } } \left( \left| \begin{array}{cccc} a_{ k , \, N-2 } & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ 0 & a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| + \left| \begin{array}{cccc} 0 & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-2 , \, N-2 } & a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \right) \\[10pt] &= \dfrac{ \left( -1 \right)^{3} }{ \prod\limits_{\nu = 1}^{ 4 } a_{ N-3+\nu , \, N-3+\nu } } \left| \begin{array}{cccc} a_{ k , \, N-2 } & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-2 , \, N-2 } & a_{ N-2 , \, N-1 } & a_{ N-2 , \, N } & a_{ N-2 , \, N+1 } \\ 0 & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ 0 & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \ \boldsymbol{.} \end{aligned} \end{align}
   ${}$
   次に，各 $r \in \left\{ \, 3 ,\, \dots ,\, N-k \, \right\}$ に対して，等式
   ${}$
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N-r+1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } = \dfrac{ \left( -1 \right)^{r} }{ \prod\limits_{\nu = 1}^{ r + 1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{ccccccc} a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & a_{ k , \, N-r+3 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & a_{ N-r+1 , \, N-r+3 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & a_{ N-r+2 , \, N-r+2 } & a_{ N-r+2 , \, N-r+3 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ 0 & 0 & a_{ N-r+3 , \, N-r+3 } & \cdots & a_{ N-r+3 , \, N-1 } & a_{ N-r+3 , \, N } & a_{ N-r+3 , \, N+1 } \\ {} & {} & \ddots & {} & \vdots & \vdots & \vdots \\ {} & {} & {} & \ddots & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \end{aligned} \tag*{\textsf{[A]}} \end{align}
   が成り立つと仮定する。このとき，
   \begin{align} \qquad \begin{aligned} a_{ k , \, N-r } \ a'_{ N-r, \, N+1 } &= a_{ k , \, N-r } \cdot \dfrac{ \left( -1 \right)^{r+1} }{ \prod\limits_{\nu = 0}^{ r+1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{cccccc} a_{ N-r , \, N-r+1 } & a_{ N-r , \, N-r+2 } & \cdots & a_{ N-r , \, N-1 } & a_{ N-r , \, N } & a_{ N-r , \, N+1 } \\ a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \\[10pt] &= \dfrac{ \left( -1 \right)^{r+1} }{ \prod\limits_{\nu = 0}^{ r+1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{ccccccc} a_{ k , \, N-r } & a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ 0 & a_{ N-r , \, N-r+1 } & a_{ N-r , \, N-r+2 } & \cdots & a_{ N-r , \, N-1 } & a_{ N-r , \, N } & a_{ N-r , \, N+1 } \\ 0 & a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \ \boldsymbol{,} \end{aligned} \end{align}
   ${}$
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N-r+1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= \dfrac{ \left( -1 \right)^{r} }{ \prod\limits_{\nu = 1}^{ r + 1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{ccccccc} a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \\[10pt] &= \left( - a_{N-r ,\, N-r} \right) \cdot \dfrac{ \left( -1 \right)^{r+1} }{ \prod\limits_{\nu = 0}^{ r + 1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{ccccccc} a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \\[10pt] &= \dfrac{ \left( -1 \right)^{r+1} }{ \prod\limits_{\nu = 0}^{ r + 1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{cccccccc} 0 & a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{N-r ,\, N-r} & a_{N-r ,\, N-r+1} & a_{N-r ,\, N-r+2} & \cdots & a_{N-r ,\, N-1} & a_{N-r ,\, N} & a_{N-r ,\, N+1} \\ 0 & a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \end{aligned} \end{align}
   となるので，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = N-r}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ k , \, N-r } \ a'_{ N-r, \, N+1 } + \sum_{\mu = N-r+1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } \\[10pt] &= \dfrac{ \left( -1 \right)^{r+1} }{ \prod\limits_{\nu = 0}^{ r+1 } a_{ N-r+\nu , \, N-r+\nu } } \left| \begin{array}{ccccccc} a_{ k , \, N-r } & a_{ k , \, N-r+1 } & a_{ k , \, N-r+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ N-r , \, N-r } & a_{ N-r , \, N-r+1 } & a_{ N-r , \, N-r+2 } & \cdots & a_{ N-r , \, N-1 } & a_{ N-r , \, N } & a_{ N-r , \, N+1 } \\ 0 & a_{ N-r+1 , \, N-r+1 } & a_{ N-r+1 , \, N-r+2 } & \cdots & a_{ N-r+1 , \, N-1 } & a_{ N-r+1 , \, N } & a_{ N-r+1 , \, N+1 } \\ 0 & 0 & a_{ N-r+2 , \, N-r+2 } & \cdots & a_{ N-r+2 , \, N-1 } & a_{ N-r+2 , \, N } & a_{ N-r+2 , \, N+1 } \\ {} & {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge{\boldsymbol{0}}} & {} & {} & {} & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \end{array} \right| \ \boldsymbol{.} \end{aligned} \end{align}
   ${}$
   ゆえに，等式 $\textsf{[A]}$ が成り立つことから，$r = N - k$ を代入した式を考えると，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= a_{ k , \, k } \ a'_{ k , \, N+1 } + \sum_{\mu = k+1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } \ \boldsymbol{,} \end{aligned} \end{align}
   ${}$
   \begin{align} \qquad \begin{aligned} a_{ k , \, k } \ a'_{ k , \, N+1 } &= a_{ k , \, k } \cdot \dfrac{ \left( -1 \right)^{ N + 1 - k } }{ \prod\limits_{\nu=0}^{ N + 1 - k } a_{ k + \nu , \, k + \nu } } \left| \begin{array}{cccccc} a_{ k , \, k+1 } & a_{ k , \, k+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ k+1 , \, k+1 } & a_{ k+1 , \, k+2 } & \cdots & a_{ k+1 , \, N-1 } & a_{ k+1 , \, N } & a_{ k+1 , \, N+1 } \\ 0 & a_{ k+2 , \, k+2 } & \cdots & a_{ k+2 , \, N-1 } & a_{ k+2 , \, N } & a_{ k+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge\boldsymbol{0}} & & & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= \dfrac{ \left( -1 \right)^{ N + 1 - k } }{ \prod\limits_{\nu=1}^{ N + 1 - k } a_{ k + \nu , \, k + \nu } } \left| \begin{array}{cccccc} a_{ k , \, k+1 } & a_{ k , \, k+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ k+1 , \, k+1 } & a_{ k+1 , \, k+2 } & \cdots & a_{ k+1 , \, N-1 } & a_{ k+1 , \, N } & a_{ k+1 , \, N+1 } \\ 0 & a_{ k+2 , \, k+2 } & \cdots & a_{ k+2 , \, N-1 } & a_{ k+2 , \, N } & a_{ k+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge\boldsymbol{0}} & & & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \ \boldsymbol{,} \end{aligned} \end{align}
   ${}$
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k+1}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } &= \dfrac{ \left( -1 \right)^{N-k} }{ \prod\limits_{\nu = 1}^{ N-k+1 } a_{ k+\nu , \, k+\nu } } \left| \begin{array}{cccccc} a_{ k , \, k+1 } & a_{ k , \, k+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ k+1 , \, k+1 } & a_{ k+1 , \, k+2 } & \cdots & a_{ k+1 , \, N-1 } & a_{ k+1 , \, N } & a_{ k+1 , \, N+1 } \\ 0 & a_{ k+2 , \, k+2 } & \cdots & a_{ k+2 , \, N-1 } & a_{ k+2 , \, N } & a_{ k+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge\boldsymbol{0}} & & & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \\[10pt] &= - \dfrac{ \left( -1 \right)^{N+1-k} }{ \prod\limits_{\nu = 1}^{ N+1-k } a_{ k+\nu , \, k+\nu } } \left| \begin{array}{cccccc} a_{ k , \, k+1 } & a_{ k , \, k+2 } & \cdots & a_{ k , \, N-1 } & a_{ k , \, N } & a_{ k , \, N+1 } \\ a_{ k+1 , \, k+1 } & a_{ k+1 , \, k+2 } & \cdots & a_{ k+1 , \, N-1 } & a_{ k+1 , \, N } & a_{ k+1 , \, N+1 } \\ 0 & a_{ k+2 , \, k+2 } & \cdots & a_{ k+2 , \, N-1 } & a_{ k+2 , \, N } & a_{ k+2 , \, N+1 } \\ {} & {} & \ddots & \vdots & \vdots & \vdots \\ {} & {} & {} & a_{ N-1 , \, N-1 } & a_{ N-1 , \, N } & a_{ N-1 , \, N+1 } \\ {\Huge\boldsymbol{0}} & & & 0 & a_{ N , \, N } & a_{ N , \, N+1 } \\ \end{array} \right| \ \boldsymbol{.} \end{aligned} \end{align}
   よって，
   \begin{align} \qquad \begin{aligned} \sum_{\mu = k}^{N+1} a_{ k , \, \mu } \ a'_{ \mu , \, N+1 } = 0 \, \boldsymbol{.} \end{aligned} \end{align}

${}$

したがって，以上のことから，$A_{N+1} A'_{N+1} = E_{N+1}$ となるので，
\begin{align} \qquad \begin{aligned} A_{n} A'_{n} &= E_{n} \, \boldsymbol{,} \end{aligned} \end{align}
すなわち，
\begin{align} \qquad \begin{aligned} A'_{n} &= A_{n}^{-1} \, \boldsymbol{.} \end{aligned} \end{align}

${}$

公式

以上の計算から，次の公式が成り立つ。

コメント

計算の煩雑さもさることながら，公式の形そのものが難しいですね。
ちなみにこの公式は，現実的な問題処理 ($n$が大きい場合) では計算量が膨大になって使い物になりません。
疎行列あたりでふわふわと計算するか，$n \leq 5$ ぐらいで計算演習に利用することをオススメします。