行列式余因子展開 4次正方行列、連立1次方程式、解、クラメルの公式

学習環境

手を動かしてまなぶ線形代数 (藤岡敦(著)、裳華房)の第3章(行列式)、9(余因子展開)、基本問題の問9.4(数物系)の解答を求めてみる。

\begin{array}{l} \det A \\ = \det [\begin{matrix} a & - b & - c & - d \\ b & a & - d & c \\ c & d & a & - b \\ d & - c & b & a \end{matrix}] \\ = {(- 1)}^{1 + 1} a \det [\begin{matrix} a & - d & c \\ d & a & - b \\ - c & b & a \end{matrix}] \\ + {(- 1)}^{1 + 2} (- b) \det [\begin{matrix} b & - d & c \\ c & a & - b \\ d & b & a \end{matrix}] \\ + {(- 1)}^{1 + 3} (- c) \det [\begin{matrix} b & a & c \\ c & d & - b \\ d & - c & a \end{matrix}] \\ + {(- 1)}^{1 + 4} (- d) \det [\begin{matrix} b & a & - d \\ c & d & a \\ d & - c & b \end{matrix}] \\ = a ((a^{3} - b c d + b c d) - (- a b^{2} - a d^{2} - a c^{2})) \\ + b ((a^{2} b + b d^{2} + b c^{2}) - (- b^{3} - a c d + a c d)) \\ - c ((a b d - a b d - c^{3}) - (b^{2} c + a^{2} c + c d^{2})) \\ + d ((b^{2} d + a^{2} d + c^{2} d) - (- a b c + a b c - d^{3})) \\ = a^{4} + a^{2} b^{2} + a^{2} d^{2} + a^{2} c^{2} \\ + a^{2} b^{2} + b^{2} d^{2} + b^{2} c^{2} + b^{4} \\ + c^{4} + b^{2} c^{2} + a^{2} c^{2} + c^{2} d^{2} \\ + b^{2} d^{2} + a^{2} d^{2} + c^{2} d^{2} + d^{4} \\ = a^{4} + 2 (b^{2} + c^{2} + d^{2}) a^{2} + 2 (b^{2} c^{2} + b^{2} d^{2} + c^{2} d^{2}) + b^{4} + c^{4} + d^{4} \\ = a^{4} + 2 (b^{2} + c^{2} + d^{2}) a^{2} + b^{4} + 2 (c^{2} + d^{2}) b^{2} + c^{4} + 2 c^{2} d^{2} + d^{4} \\ = a^{4} + 2 (b^{2} + c^{2} + d^{2}) a^{2} + b^{4} + 2 (c^{2} + d^{2}) b^{2} + {(c^{2} + d^{2})}^{2} \\ = a^{2} + 2 (b^{2} + c^{2} + d^{2}) a^{2} + {(b^{2} + c^{2} + dz)}^{2} \\ = {(a^{2} + b^{2} + c^{2} + d^{2})}^{2} \end{array}

x = (x_{1}, x_{2}, x_{3}, x_{4})

とおく。

\begin{array}{l} x_{1} = \frac{1}{\det A} [\begin{matrix} 1 & - b & - c & - d \\ 0 & a & - d & c \\ 0 & d & a & - b \\ 0 & - c & b & a \end{matrix}] \\ = \frac{1}{{(a^{2} + b^{2} + c^{2} + d^{2})}^{2}} (a^{3} + a b^{2} + a c^{2} + a d^{2}) \\ = \frac{a}{a^{2} + b^{2} + c^{2} + d^{2}} \end{array}

同様に計算して、

x = \frac{1}{a^{2} + b^{2} + c^{2} + d^{2}} [\begin{matrix} a \\ b \\ c \\ d \end{matrix}]