[理工] 矩陣 問題
題目是這樣
| a 1/3 2/3 |
| |
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A=| b 2/3 1/3 | 為正交矩陣 求abc值
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| c 2/3 -2/3 |
T
解:因為A為正交矩陣 A A=I ,令A={A1 A2 A3}
T
[A1 A2 A3] [A1 A2 A3] = I
T 2 2 2
| a b c | { A1 A1 =a + b + c = 1 << "1沒錯的話是I"
| | { T
T | | 可得{ A2 A1 =1/3(a+2b+2c) = 0 <<怎麼會等於"零"
A =| 1/3 2/3 2/3 | , { T
| | { A3 A1 =1/3(2A+B-2C) = 0 <<怎麼會等於零....
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| 2/3 1/3 -2/3 |
幫補血加速天賜霸邪>.<|||||
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