{"id":425,"date":"2012-06-14T23:41:29","date_gmt":"2012-06-14T14:41:29","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=425"},"modified":"2021-10-20T14:54:34","modified_gmt":"2021-10-20T05:54:34","slug":"ykr201202","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/ykr201202\/","title":{"rendered":"\u6a2a\u56fd\u5927\u7406\u7cfb2012\uff1a\u7b2c2\u554f"},"content":{"rendered":"
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\u884c\u5217 \\(X =\\left( \\begin{array}{cc} a & b \\\\ c & d \\end{array} \\right) , \\, Y =\\left( \\begin{array}{cc} -2 & 3 \\\\ 3 & 6 \\end{array} \\right)\\) \u306f \\(XY=YX\\) \u3092\u6e80\u305f\u3059.\r\n\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(c ,d\\) \u3092 \\(a , b\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\\(X^2=E , \\, b \\gt 0\\) \u306e\u3068\u304d, \\(X\\) \u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \\(E\\) \u306f\u5358\u4f4d\u884c\u5217\u3068\u3059\u308b.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\\(xy\\) \u5e73\u9762\u4e0a\u306b\u76f4\u7dda \\(\\ell\\) \u304c\u3042\u308a, (2)<\/strong> \u3067\u6c42\u3081\u305f\u884c\u5217 \\(X\\) \u306e\u8868\u3059 \\(1\\) \u6b21\u5909\u63db\u306b\u3088\u3063\u3066 \\(\\ell\\) \u4e0a\u306e\u70b9\u306f\u3059\u3079\u3066 \\(\\ell\\) \u4e0a\u306e\u70b9\u306b\u79fb\u3055\u308c\u308b. \\(\\ell\\) \u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n

    (1)<\/strong><\/p>\r\n

    \\[\\begin{align}\r\nXY & = \\left( \\begin{array}{cc} -2a+3b & 3a+6b \\\\ -2c+3d & 3c+6d \\end{array} \\right) \\ , \\\\\r\nYX & = \\left( \\begin{array}{cc} -2a+3b & -2b+3d \\\\ 3a+6c & 3c+6d \\end{array} \\right)\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\((1,1)\\) \u6210\u5206\u3092\u6bd4\u8f03\u3057\u3066\r\n\\[\\begin{align}\r\n-2a+3b & = -2a+3c \\\\\r\n\\text{\u2234} \\quad c & = \\underline{b}\r\n\\end{align}\\]\r\n\\((1,2)\\) \u6210\u5206\u3092\u6bd4\u8f03\u3057\u3066\r\n\\[\\begin{align}\r\n3a+6b & = -2b+3d \\\\\r\n\\text{\u2234} \\quad d & = \\underline{a+\\dfrac{8b}{3}}\r\n\\end{align}\\]\r\n

    (2)<\/strong><\/p>\r\n

    \u30b1\u30fc\u30ea\u30fc\u30fb\u30cf\u30df\u30eb\u30c8\u30f3\u306e\u5b9a\u7406\u3088\u308a\r\n\\[\r\na+d=0 \\ , \\ ad-bc=-1 \\quad ... [1]\r\n\\]\r\n[1] \u3068 (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\na+d & = 2a +\\dfrac{8b}{3} = 0 \\\\\r\n\\text{\u2234} \\quad a & = -\\dfrac{4b}{3} \\\\\r\n\\text{\u2234} \\quad d & = a+\\dfrac{8b}{3} = \\dfrac{4b}{3}\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nad-bc & = \\left( -\\dfrac{4b}{3} \\right) \\cdot \\dfrac{4b}{3} -b^2 = -1 \\\\\r\n\\dfrac{25 b^2}{9} & = 1 \\\\\r\n\\text{\u2234} \\quad b & = \\dfrac{3}{5} \\quad ( \\ \\text{\u2235} \\ b \\gt 0 \\ )\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nX =\\underline{\\left( \\begin{array}{cc} -\\dfrac{4}{5} & \\dfrac{3}{5} \\\\ \\dfrac{3}{5} & \\dfrac{4}{5} \\end{array} \\right)}\r\n\\]\r\n

    (3)<\/strong><\/p>\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(\\ell : \\ x = a\\) \u3068\u8868\u3055\u308c\u308b\u3068\u304d
      \r\n\\(\\ell\\) \u4e0a\u306e\u70b9\u306f \\((a,t)\\) \u3068\u8868\u305b\u3066\r\n\\[\r\nX \\left( \\begin{array}{c} a \\\\ t \\end{array} \\right) = \\dfrac{1}{5} \\left( \\begin{array}{c} -4a+3t \\\\ 3a+4t \\end{array} \\right)\r\n\\]\r\n\u3053\u306e\u70b9\u304c \\(\\ell\\) \u4e0a\u306b\u3042\u308b\u306e\u3067\r\n\\[\r\n\\dfrac{-4a+3t}{5}= a\r\n\\]\r\n\u3053\u308c\u304c\u4efb\u610f\u306e \\(t\\) \u306b\u3064\u3044\u3066\u6210\u7acb\u3059\u308b\u3053\u3068\u306f\u306a\u3044\u306e\u3067, \u4e0d\u9069.<\/p><\/li>\r\n

    2. 2*<\/strong>\u3000\\(\\ell : \\ y = ax+b\\) \u3068\u8868\u3055\u308c\u308b\u3068\u304d
      \r\n\\(\\ell\\) \u4e0a\u306e\u70b9\u306f \\(( t , at+b )\\) \u3068\u8868\u305b\u3066\r\n\\[\\begin{align}\r\nX \\left( \\begin{array}{c} t \\\\ at+b \\end{array} \\right) & = \\dfrac{1}{5} \\left( \\begin{array}{c} -4t+3at+3b \\\\ 3t+4at+4b \\end{array} \\right) \\\\\r\n& = \\dfrac{1}{5} \\left( \\begin{array}{c} (3a-4)t+3b \\\\ (4a+3)t+4b \\end{array} \\right)\r\n\\end{align}\\]\r\n\u3053\u306e\u70b9\u304c \\(\\ell\\) \u4e0a\u306b\u3042\u308b\u306e\u3067\r\n\\[\\begin{align}\r\n\\dfrac{(4a+3)t+4b}{5} = a \\cdot \\dfrac{(3a-4)t+3b}{5} & +b \\\\\r\n(3a^2-8a-3)t +3ab+b & = 0 \\\\\r\n\\text{\u2234} \\quad (3a+1)(3a-1)t +(3a+1)b & = 0\r\n\\end{align}\\]\r\n\u3053\u308c\u304c\u4efb\u610f\u306e \\(t\\) \u306b\u3064\u3044\u3066\u6210\u7acb\u3059\u308b\u306e\u3067\r\n\\[\\begin{align}\r\n(3a+1)(3a-1) & = 0 \\quad ... [2] \\ , \\\\\r\n(3a+1)b & = 0 \\quad ... [3]\r\n\\end{align}\\]\r\n[2] \u3088\u308a\r\n\\[\r\na = \\pm \\dfrac{1}{3}\r\n\\]\r\n[3] \u3088\u308a\r\n