2次の行列と行列式累乗、逆行列、非零、等式

学習環境

行列と行列式 (現代数学への入門) (砂田利一(著)、岩波書店)の第1章(2次の行列と行列式)、演習問題1.8の解答を求めてみる。

\begin{array}{l} A = M^{2} + 3 M + 2 I \\ = [\begin{matrix} a & 1 \\ a & a \end{matrix}] [\begin{matrix} a & 1 \\ a & a \end{matrix}] + [\begin{matrix} 3 a & 3 \\ 3 a & 3 a \end{matrix}] + [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \\ = [\begin{matrix} a^{2} + a & 2 a \\ 2 a^{2} & a^{2} + a \end{matrix}] + [\begin{matrix} 3 a & 3 \\ 3 a & 3 a \end{matrix}] + [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \\ = [\begin{matrix} a^{2} + 4 a + 2 & 2 a + 3 \\ 2 a^{2} + 3 a & a^{2} + 4 a + 2 \end{matrix}] \end{array}

\begin{array}{l} A = M^{2} + 3 M + 2 I \\ A = (M + 2 I) (M + I) \end{array}

よって、

\begin{array}{l} B = M + I \\ = [\begin{matrix} a + 1 & 1 \\ a & a + 1 \end{matrix}] \end{array}

\begin{array}{l} \det A \\ = \det ((M + 2 I) B) \\ = \det (M + 2 I) \det B \\ = \det [\begin{matrix} a + 2 & 1 \\ a & a + 2 \end{matrix}] \det [\begin{matrix} a + 1 & 1 \\ a & a + 1 \end{matrix}] \\ = ({(a + 2)}^{2} - a) ({(a + 1)}^{2} - a) \\ = (a^{2} + 3 a + 4) (a^{2} + a + 1) \\ = ({(a + \frac{3}{2})}^{2} + \frac{7}{4}) ({(a + \frac{1}{2})}^{2} + \frac{3}{4}) \\ > 0 \end{array}

よってAは逆行列をもつ。

その逆行列を求める。

\begin{array}{l} A^{- 1} \\ = \frac{1}{(a^{2} + 3 a + 4) (a^{2} + a + 1)} [\begin{matrix} a^{2} + 4 a + 2 & - (2 a + 3) \\ - (2 a^{2} + 3 a) & a^{2} + 4 a + 2 \end{matrix}] \end{array}

\begin{array}{l} 2 a + 3 = 0 \\ a = - \frac{3}{2} \\ p = q \\ = \frac{a^{2} + 4 a + 2}{(a^{2} + 3 a + 4) (a^{2} + a + 1)} \\ = \frac{\frac{9}{4} - 6 + 2}{(\frac{9}{4} - \frac{9}{2} + 4) (\frac{9}{4} - \frac{3}{2} + 1)} \\ = \frac{9 - 24 + 8}{4} \cdot \frac{4}{9 - 18 + 16} \cdot \frac{4}{9 - 6 + 4} \\ = \frac{- 7 \cdot 4}{7 \cdot 7} \\ = - \frac{4}{7} \end{array}