\\(a\\) \u3092\u5b9f\u6570\u3068\u3057\u3066, \\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217 \\(A , B\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u3081\u308b.\r\n\\[\r\nA = \\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)\r\n\\]\r\n\u3053\u306e\u3068\u304d, \\(\\left\\{ ( \\cos t ) A +( \\sin t ) B \\right\\}^2 = O\\) \u3092\u307f\u305f\u3059\u5b9f\u6570 \\(t\\) \u304c\u5b58\u5728\u3059\u308b\u3088\u3046\u306a \\(a\\) \u306e\u7bc4\u56f2\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \\(O\\) \u306f\u96f6\u884c\u5217\u3068\u3059\u308b.<\/p>\r\n
\\[\\begin{align}\r\nA^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) \\\\\r\n& = \\left( \\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} \\right) = E , \\\\\r\nB^2 & = \\left( \\begin{array}{cc} a & 0 \\\\ 2 & -a \\end{array} \\right) \\left( \\begin{array}{cc} a & 0 \\\\ 2 & -a \\end{array} \\right) \\\\\r\n& = \\left( \\begin{array}{cc} a^2 & 0 \\\\ 0 & a^2 \\end{array} \\right) = a^2 E , \\\\\r\nAB & = \\left( \\begin{array}{cc} 1 & a+1 \\\\ 0 & -1 \\end{array} \\right) \\left( \\begin{array}{cc} a & 0 \\\\ 2 & -a \\end{array} \\right) \\\\\r\n& = \\left( \\begin{array}{cc} 4a+2 & -a^2-a \\\\ -2 & a \\end{array} \\right) , \\\\\r\nBA & = \\left( \\begin{array}{cc} a & 0 \\\\ 2 & -a \\end{array} \\right) \\left( \\begin{array}{cc} 1 & a+1 \\\\ 0 & -1 \\end{array} \\right) \\\\\r\n& = \\left( \\begin{array}{cc} 0 & a^2+a \\\\ 2 & 3a+2 \\end{array} \\right)\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nAB +BA & = \\left( \\begin{array}{cc} 4a+2 & 0 \\\\ 0 & 4a+2 \\end{array} \\right) \\\\\r\n& = 2 (2a+1) E\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{gather}\r\n\\left\\{ ( \\cos t ) A +( \\sin t ) B \\right\\}^2 = O \\\\\r\n\\left\\{ \\cos^2 t +2( 2a+1 ) \\sin t \\cos t +a^2 \\sin^2 t \\right\\} E = O\r\n\\end{gather}\\]\r\n\u3086\u3048\u306b\r\n\\[\\begin{gather}\r\n\\cos^2 t +2( 2a+1 ) \\sin t \\cos t +a^2 \\sin^2 t = 0 \\\\\r\n\\dfrac{1+\\cos 2t}{2} +( 2a+1 ) \\sin 2t +\\dfrac{a^2 ( 1-\\cos 2t )}{2} = 0 \\\\\r\n2 ( 2a+1 ) \\sin 2t +( a^2-1 ) \\cos 2t +( a^2+1 ) = 0 \\\\\r\n\\sqrt{4(2a+1)^2+(a^2-1)^2} \\sin ( 2t +\\alpha ) = -(a^2+1) \\\\\r\n\\text{\u2234} \\quad \\sin ( 2t +\\alpha ) = -\\dfrac{a^2+1}{\\sqrt{4(2a+1)^2+(a^2-1)^2}}\r\n\\end{gather}\\]\r\n\u3053\u308c\u3092\u307f\u305f\u3059 \\(t\\) \u304c\u5b58\u5728\u3059\u308b\u6761\u4ef6\u306f\r\n\\[\\begin{gather}\r\n\\dfrac{a^2+1}{\\sqrt{4(2a+1)^2+(a^2-1)^2}} \\leqq 1 \\\\\r\n(a^2+1)^2 \\leqq 4(2a+1)^2+(a^2-1)^2 \\\\\r\n3a^2+4a+1 \\geqq 0 \\\\\r\n(3a+1)(a+1) \\geqq 0 \\\\\r\n\\text{\u2234} \\quad \\underline{a \\leqq -1 , -\\dfrac{1}{3} \\leqq a}\r\n\\end{gather}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a\\) \u3092\u5b9f\u6570\u3068\u3057\u3066, \\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217 \\(A , B\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u3081\u308b. \\[ A = \\left( \\begin{array}{cc} 1 & a+1 \\\\ 0 & -1 \\end{array} … \u7d9a\u304d\u3092\u8aad\u3080