行列式連立1次方程式と行列式 2次行列式と3次行列式の計算

学習環境

線形代数演習〈理工系の数学入門コース/演習新装版〉 (浅野功義(著)、大関清太(著)、岩波書店)の第3章(行列式)、3-1(連立1次方程式と行列式)、問題1、2の解答を求めてみる。

1-1

\det [\begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix}] = 8 - 6 = 2

1-2

\det [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] = \cos^{2} θ + \sin^{2} θ = 1

1-3

\det [\begin{matrix} 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3} \end{matrix}] = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}

1-4

\det [\begin{matrix} a + b & a + 3 b \\ a + 2 b & a + 4 b \end{matrix}]

= (a^{2} + 5 b + 4 b^{2}) - (a^{2} + 5 b + 6 b^{2})

= - 2 b^{2}

2-1

\det [\begin{matrix} - 1 & 0 & 3 \\ 2 & 1 & 5 \\ 2 & - 3 & 4 \end{matrix}]

= (- 4 - 18) - (15 + 6)

= - 22 - 21

= - 43

2-2

\det [\begin{matrix} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{matrix}]

= (\frac{1}{3} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{4}) - (\frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{5} + \frac{1}{3^{3}})

= \frac{1}{3 \cdot 5} + \frac{1}{2^{2} \cdot 3} - \frac{1}{2^{4}} - \frac{1}{2^{2} \cdot 5} + \frac{1}{3^{3}}

= \frac{2^{4} \cdot 3^{2} + 2^{2} \cdot 3^{2} \cdot 5 - 3^{3} \cdot 5 - 2^{2} \cdot 3^{3} - 2^{4} \cdot 5}{2^{4} \cdot 3^{3} \cdot 5}

= \frac{144 + 180 - 135 - 108 - 80}{2^{4} \cdot 3^{3} \cdot 5}

= \frac{1}{2160}

2-3

\det [\begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix}]

= a^{3} + b^{3} + c^{3} - 3 a b c

= (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)