※ 引述《mqazz1 (無法顯示)》之銘言： : 已知 a(1)a(4) = a(2)a(3) : [√[a(1)a(4)] 0 0 a(1) ] : [ 0 √[a(1)a(4)] a(2) 0 ] : [ 0 a(3) √[a(1)a(4)] 0 ] : [ a(4) 0 0 √[a(1)a(4)]] : 第1列和第4列為linearly dependent : 第2列和第3列為linearly dependent : 書上寫 : ([√[a(1)a(4)] a(2) ]) : det([ a(3) √[a(1)a(4)]]) = 0 : ([√[a(1)a(4)] a(1) ]) : det([ a(4) √[a(1)a(4)]]) = 0 : 請問為什麼可以這樣判斷? : 謝謝 因為，如果第一列跟第四列線性相依(linearly depedent) 的話， ＿＿＿＿＿ ＿＿＿＿＿ α[ √[a(1)a(4)] 0 0 a(1) ] ＝ β[ a(4) 0 0 √[a(1)a(4)] ] ＿＿＿＿＿ ＿＿＿＿＿ <=> α[ √[a(1)a(4)] a(1) ] ＝ β[ a(4) √[a(1)a(4)] ] ＿＿＿＿＿ ╭ √[a(1)a(4)] a(1) ╮ <=> det│ ＿＿＿＿＿ │＝ 0 for some α,β 在體 |F. ╰ a(4) √[a(1)a(4)] ╯ 同理， 由第二列和第三列線性相依可得 ＿＿＿＿＿ ╭ √[a(1)a(4)] a(2) ╮ det│ ＿＿＿＿＿ │ ╰ a(3) √[a(1)a(4)] ╯ 再看不懂的話，用降階直接算(但也沒全部算就是了) ＿＿＿＿＿ ╭ √[a(1)a(4)] 0 0 a(1) ╮ │ 0 √[a(1)a(4)] a(2) 0 │ det│ 0 a(3) √[a(1)a(4)] 0 │ ╰ a(4) 0 0 √[a(1)a(4)] ╯ ＿＿＿＿＿ ＿＿＿＿＿ ╭ √[a(1)a(4)] a(2) 0 ╮ ＝ √[a(1)a(4)] det│ a(3) √[a(1)a(4)] 0 │ ╰ 0 0 √[a(1)a(4)] ╯ ＿＿＿＿＿ ╭ 0 √[a(1)a(4)] a(2) ╮ － a(1) det│ 0 a(3) √[a(1)a(4)] │ ╰ a(4) 0 0 ╯ ＿＿＿＿＿ ＿＿＿＿＿ 2 ╭ √[a(1)a(4)] a(2) ╮ ＝ (√[a(1)a(4)]) det│ ＿＿＿＿＿ │ ╰ a(3) √[a(1)a(4)] ╯ ＿＿＿＿＿ ╭ √[a(1)a(4)] a(2) ╮ － a(1)a(4) det│ ＿＿＿＿＿ │. ╰ a(3) √[a(1)a(4)] ╯ ＿＿＿＿＿ ╭ √[a(1)a(4)] a(2) ╮ ＿＿＿＿＿ 2 Since det│ ＿＿＿＿＿ │＝ (√[a(1)a(4)]) － a(2)a(3) ╰ a(3) √[a(1)a(4)] ╯ ＝ a(1)a(4) － a(2)a(3) ＝ 0, the determinant is zero. -- 發動魔法卡「交換」! ╲ ◢ ∕ ╱ ◤▆▄▅▆ █ ﹨ ≡ 從我手牌中抽張牌吧! ╲ ◢ ∕ ╱ ◤ ▃▅▅▆▆ ◤▆▃▁ 原圖ψbloodlime ◥ ╲ ▃▆ ∕╱ ▼ ◣ ◢ █◣█ ∕ ﹨ http://ppt.cc/c425 ▃▄█▂▂▃∕ ﹨∥◢◥█ ╲﹨∕╱ ▌ █◤ █∕ ﹨ http://ppt.cc/~Mea ╲ ▼∥﹨∥◣ █◢ █◣ █◢▋ ╱ ▼╱ http://ppt.cc/IV;8 ←三選一 ╲ ▼∥﹨ ████ ﹨ █████ ╱ ▼ ￣ --

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