\\(a \\neq 0\\) \u3068\u3059\u308b.\r\n\\(A = \\left( \\begin{array}{c} 0 \\\\ 1 \\end{array} \\right)\\) , \\(B = \\left( \\begin{array}{c} a \\\\ b \\end{array} \\right)\\) \u306b\u5bfe\u3057\u3066, \\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217 \\(P\\) \u304c\r\n\\[\r\nPA = A+B , \\ PB = A-B\r\n\\]\r\n\u3092\u6e80\u305f\u3057\u3066\u3044\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(a\\) \u3068 \\(b\\) \u3092\u7528\u3044\u3066 \\(P\\) \u3092\u8868\u3057, \\(P\\) \u304c\u9006\u884c\u5217 \\(P^{-1}\\) \u3092\u3082\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(\\left( \\begin{array}{c} s \\\\ t \\end{array} \\right) = P^{-1} B\\) \u3068\u304a\u304f. \\(xy\\) \u5e73\u9762\u306b\u304a\u3044\u3066, \u70b9 \\((a,b)\\) \u304c\u653e\u7269\u7dda \\(x = y^2+2\\) \u4e0a\u3092\u52d5\u304f\u3068\u304d, \u70b9 \\((s,t)\\) \u306e\u8ecc\u8de1\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \\(|y| \\leqq 1\\) \u3068\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6761\u4ef6\u3088\u308a\r\n\\[\r\nP \\left( \\begin{array}{cc} 0 & a \\\\ 1 & b \\end{array} \\right) = \\left( \\begin{array}{cc} a & -a \\\\ 1+b & 1-b \\end{array} \\right) \\quad ... [1]\r\n\\]\r\n\u3053\u3053\u3067\r\n\\[\r\n\\det \\left( \\begin{array}{cc} 0 & a \\\\ 1 & b \\end{array} \\right) = -a \\neq 0\r\n\\]\r\n\u306a\u306e\u3067, \\(\\left( \\begin{array}{cc} 0 & a \\\\ 1 & b \\end{array} \\right)^{-1}\\) \u304c\u5b58\u5728\u3059\u308b\u306e\u3067, [1]\u306e\u4e21\u8fba\u306b\u53f3\u304b\u3089\u639b\u3051\u3066\r\n\\[\\begin{align}\r\nP & = -\\dfrac{1}{a} \\left( \\begin{array}{cc} a & -a \\\\ 1+b & 1-b \\end{array} \\right) \\left( \\begin{array}{cc} b & -a \\\\ -1 & 0 \\end{array} \\right) \\\\\r\n& = \\underline{\\left( \\begin{array}{cc} -b-1 & a \\\\ -\\dfrac{b^2+2b-1}{a} & b+1 \\end{array} \\right)}\r\n\\end{align}\\]\r\n\u3053\u306e\u3068\u304d\r\n\\[\\begin{align}\r\n\\det P & = -(b+1)^2 +(b^2+2b-1) \\\\\r\n& = -2 \\neq 0\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \u9006\u884c\u5217 \\(P^{-1}\\) \u306f\u5b58\u5728\u3059\u308b.<\/p>\r\n (2)<\/strong><\/p>\r\n \u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\nP(A-B) & = PA-PB \\\\\r\n& = (A+B) -(A-B) = 2B \\\\\r\n\\text{\u2234} \\quad P^{-1} B & = \\dfrac{1}{2} (A-B)\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\n\\left( \\begin{array}{c} s \\\\ t \\end{array} \\right) & = \\dfrac{1}{2} \\left( \\begin{array}{c} -a \\\\ 1-b \\end{array} \\right) \\\\\r\n\\text{\u2234} \\quad a = -2s & , \\ b = 1-2t\r\n\\end{align}\\]\r\n\u3053\u308c\u3092 \\(a = b^2+2\\) \u306b\u4ee3\u5165\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n-2s & = (1-2t)^2 +2 \\\\\r\n-2s & = 4t^2 -4t +3 \\\\\r\n\\text{\u2234} \\quad s & = -2 \\left( t -\\dfrac{1}{2} \\right)^2 -1\r\n\\end{align}\\]\r\n\u305f\u3060\u3057\r\n\\[\\begin{align}\r\n|b| & = |1-2t| \\leqq 1 \\\\\r\n\\text{\u2234} \\quad 0 & \\leqq t \\leqq 2\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u70b9 \\((s,t)\\) \u306e\u8ecc\u8de1\u306f\r\n\\[\r\n\\underline{\\text{\u653e\u7269\u7dda} : \\ s = -2(t-1)^2 -1 \\ \\left( \\dfrac{1}{2} \\leqq t \\leqq \\dfrac{3}{2} \\right)}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a \\neq 0\\) \u3068\u3059\u308b. \\(A = \\left( \\begin{array}{c} 0 \\\\ 1 \\end{array} \\right)\\) , \\(B = \\left( \\begin{array}{c} a … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n