\(a\) を実数として, \(2\) 次の正方行列 \(A , B\) を次のように定める. \[ A = \left( \begin{array}{cc} 1 & a+1 \\ 0 & -1 \end{array} \right) , \quad B = \left( \begin{array}{cc} a & 0 \\ 2 & -a \end{array} \right) \] このとき, \(\left\{ ( \cos t ) A +( \sin t ) B \right\}^2 = O\) をみたす実数 \(t\) が存在するような \(a\) の範囲を求めよ. ただし, \(O\) は零行列とする.

【 解 答 】

\[\begin{align} A^2 & = \left( \begin{array}{cc} 1 & a+1 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 1 & a+1 \\ 0 & -1 \end{array} \right) \\ & = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) = E , \\ B^2 & = \left( \begin{array}{cc} a & 0 \\ 2 & -a \end{array} \right) \left( \begin{array}{cc} a & 0 \\ 2 & -a \end{array} \right) \\ & = \left( \begin{array}{cc} a^2 & 0 \\ 0 & a^2 \end{array} \right) = a^2 E , \\ AB & = \left( \begin{array}{cc} 1 & a+1 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} a & 0 \\ 2 & -a \end{array} \right) \\ & = \left( \begin{array}{cc} 4a+2 & -a^2-a \\ -2 & a \end{array} \right) , \\ BA & = \left( \begin{array}{cc} a & 0 \\ 2 & -a \end{array} \right) \left( \begin{array}{cc} 1 & a+1 \\ 0 & -1 \end{array} \right) \\ & = \left( \begin{array}{cc} 0 & a^2+a \\ 2 & 3a+2 \end{array} \right) \end{align}\] なので \[\begin{align} AB +BA & = \left( \begin{array}{cc} 4a+2 & 0 \\ 0 & 4a+2 \end{array} \right) \\ & = 2 (2a+1) E \end{align}\] したがって \[\begin{gather} \left\{ ( \cos t ) A +( \sin t ) B \right\}^2 = O \\ \left\{ \cos^2 t +2( 2a+1 ) \sin t \cos t +a^2 \sin^2 t \right\} E = O \end{gather}\] ゆえに \[\begin{gather} \cos^2 t +2( 2a+1 ) \sin t \cos t +a^2 \sin^2 t = 0 \\ \dfrac{1+\cos 2t}{2} +( 2a+1 ) \sin 2t +\dfrac{a^2 ( 1-\cos 2t )}{2} = 0 \\ 2 ( 2a+1 ) \sin 2t +( a^2-1 ) \cos 2t +( a^2+1 ) = 0 \\ \sqrt{4(2a+1)^2+(a^2-1)^2} \sin ( 2t +\alpha ) = -(a^2+1) \\ \text{∴} \quad \sin ( 2t +\alpha ) = -\dfrac{a^2+1}{\sqrt{4(2a+1)^2+(a^2-1)^2}} \end{gather}\] これをみたす \(t\) が存在する条件は \[\begin{gather} \dfrac{a^2+1}{\sqrt{4(2a+1)^2+(a^2-1)^2}} \leqq 1 \\ (a^2+1)^2 \leqq 4(2a+1)^2+(a^2-1)^2 \\ 3a^2+4a+1 \geqq 0 \\ (3a+1)(a+1) \geqq 0 \\ \text{∴} \quad \underline{a \leqq -1 , -\dfrac{1}{3} \leqq a} \end{gather}\]