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PinkyPYC一星新手會員
2010-12-18 18:08#1
有條數學阿Sir話要用matrices來做，但做極都唔識，身邊又冇人可以問到，請各位多多幫忙一下，在下不勝感激﹗﹗﹗:ro02:
Solve the following system of inhomogeneous linear equations:
4X3 + X4 = 1
2X1 + 2X2 – 2X3 + 5X4 = 1
5X1 + 5X2 - X3 + 5X4 = 2
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雪子秀版主
2010-12-18 20:24#2
X前面是未知數？如果是，那麼就"矩陣方程"。:smilie_O_o:
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PinkyPYC一星新手會員
2010-12-18 20:42#3
係要矩陣方程來做，不過唔識做...
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36661124二轉會員
2010-12-18 21:49#4
Solve the following system of inhomogeneous linear equations:
4X3 + X4 = 1
2X1 + 2X2 – 2X3 + 5X4 = 1
5X1 + 5X2 - X3 + 5X4 = 2
變返個MATRIX先:寫法: (打直其實係一個大括號 ie 4x4 matrix)
(0 0 4 1 )( X1) ( 1)
(2 2 -2 5 ) (X2) (1)
(5 5 -1 5 )( X3) = (2)
(0 0 0 0 )(X4) (0)
先搵
(0 0 4 1 )
(2 2 -2 5 )
(5 5 -1 5 )
(0 0 0 0 )
1 .E舊野既DET ,證左唔=0
2.之後搵INVERSE
3.搵到INVERSE (假設叫A) 之後 係個MATRIX 左 右自 X inverse in order to cancel out the matrix of
(0 0 4 1 )
(2 2 -2 5 )
(5 5 -1 5 )
(0 0 0 0 )
hence you will get
( X1) A ( 1)
(X2) (1)
( X3) = (2)
(X4) (0)
THEN THE RESULT AS FOLLOW
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NaozumiVIP會員
2010-12-19 09:35#5
樓上的朋友有點錯, 個matrix有一行全係0,咁個det = 0
其實用Gaussian elimination會好 些,搵inverse真係好易錯
先考慮增廣矩陣(augmented matrix)
(0 0 4 1 1)
(2 2 -2 5 1)
(5 5 -1 5 2)
(2 2 -2 5 1)
(5 5 -1 5 2)
(0 0 4 1 1)
(2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 4 1 1)
(2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 0 8.5 1.5)
用back substitution計到x4, x3.
x1 同 x2係 用2X1 + 2X2 – 2X3 + 5X4 = 1 relate番
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36661124二轉會員
2010-12-19 10:29#6
原帖由 Naozumi 於 10-12-19 09:35 AM 發表 
樓上的朋友有點錯, 個matrix有一行全係0,咁個det = 0
其實用Gaussian elimination會好 些,搵inverse真係好易錯
先考慮增廣矩陣(augmented matrix)
(0 0 4 1 1)
(2 2 ...
thx:smilie_:P2:
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PinkyPYC一星新手會員
2010-12-19 10:58#7
「(2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 0 8.5 1.5)」
計到呢度之後，可唔可以教埋我跟住既step係點樣計落去呀？
我真係唔係好識計...:ro03:
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36661124二轉會員
2010-12-19 12:12#8
原帖由 PinkyPYC 於 10-12-19 10:58 AM 發表
「(2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 0 8.5 1.5)」
計到呢度之後，可唔可以教埋我跟住既step係點樣計落去呀？
我真係唔係好識計 ...
8.5(x4)=1.5
x4=....
跟住將x4 sub返入去第2條式 搵到x3
將x3 x4 sub入第1條式
x1 x2 應該係varaible
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PinkyPYC一星新手會員
2010-12-19 17:06#9
(1) (2 2 -2 5 1)
(5 5 -1 5 2)
(0 0 4 1 1)
(2) (2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 4 1 1)
(3) (2 2 -2 5 1)
(0 0 4 -7.5 -0.5)
(0 0 0 8.5 1.5)
我想知(2)既row2係點來？係咪將(1)既row2除5？如果係咁計，咁(2)既row2後三個數即係點來？
同埋(3)既row3又係點得出來？
[ 本帖最後由 PinkyPYC 於 10-12-19 05:09 PM 編輯 ]
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NaozumiVIP會員
2010-12-19 18:49#10
我係將(1)的 row 1 x (-5/2) + row 2 --> row 2
你最後一個問題: 將(2)的 row 2 x (-1) + row 3 --> row 3
其實只係用係初中學過的method of elinination的方法消除variable
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PinkyPYC一星新手會員
2010-12-19 20:00#11
咁計完個ans係咩呀？係咪X1+X2=19/34，X3=7/34，X4=3/17？
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badboytim三星高級會員
2010-12-20 01:10#12
4X3 + X4 = 1
2X1 + 2X2 – 2X3 + 5X4 = 1
5X1 + 5X2 - X3 + 5X4 = 2
rewrited as :
4c +d =1
2a +2b -2c +5d =1
5a +5b -1c +5d =2
R1: 2 2 -2 5 | 1
R2: 5 5 -1 5 | 2
R3: 0 0 4 1 | 1
1. R1X0.5 --> R1
2. R2X0.2 --> R2
R1: 1 1 -1 2.5 | 0.5
R2: 1 1 -0.2 1 | 0.4
R3: 0 0 4 1 | 1
3. R2 - R1 -->R2
R1: 1 1 -1 2.5 | 0.5
R2: 0 0 0.8 -1.5 |-0.1
R3: 0 0 4 1 | 1
4. R2/0.8 -->R2
5. R3/4 -->R3
R1: 1 1 -1 2.5 | 0.5
R2: 0 0 1 -15/8 | -1/8
R3: 0 0 1 1/4 | 1/4
6. R3 -R2 -->R3
R1: 1 1 -1 2.5 | 0.5
R2: 0 0 1 -15/8 | -1/8
R3: 0 0 0 17/8 | 3/8
7. R3 x 8/17 -->R3
R1: 1 1 -1 2.5 | 0.5
R2: 0 0 1 -15/8 | -1/8
R3: 0 0 0 1 | 3/17
8. R1 - 2.5 x R3 --> R1
9. R2 + 15/8 x R3 -->R2
R1: 1 1 -1 0 | 2/34
R2: 0 0 1 0 | 7/34
R3: 0 0 0 1 | 3/17
10. R1 +R2 --> R1
R1: 1 1 0 0 | 9/34
R2: 0 0 1 0 | 7/34
R3: 0 0 0 1 | 3/17
therefore
a+b = 9/34
c = 7/34
d = 3/17
Thanks for the correction.
[ 本帖最後由 badboytim 於 10-12-20 02:35 PM 編輯 ]
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badboytim三星高級會員
2010-12-20 01:14#13
tips:
you can check your ans. by subsititue the ans. into the given equation,
and notice that the ans. above ( 1st -10th replyers' ans.) are wrong.
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NaozumiVIP會員
2010-12-20 11:46#14
badboytim兄
step 5 R2 最後的一個數應為 -1/8
而你的答案並不符合最後一條方程 5X1 + 5X2 - X3 + 5X4 = 2
[ 本帖最後由 Naozumi 於 10-12-20 11:51 AM 編輯 ]
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NaozumiVIP會員
2010-12-20 13:07#15
原帖由 PinkyPYC 於 10-12-19 20:00 發表
咁計完個ans係咩呀？係咪X1+X2=19/34，X3=7/34，X4=3/17？
x1 + x2計1錯
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PinkyPYC一星新手會員
2010-12-25 15:18#16
謝謝各位大哥出手相助﹗﹗﹗:ro02: