作者ntust661 (常春藤 i pod)
看板Math
標題Re: [工數] 矩陣問題
時間Sat Aug 11 23:21:01 2012
※ 引述《kilikolo0218 (小霸王)》之銘言:
: dX
: 考慮狀態方程式: ----- = AX
: dt
: 2 1 3
: 其中A = [ 0 2 1 ]
: 0 0 4
: 試求 exp(At)。
: 我是想說是否求出X就是答案了
: 但不確定
: 感謝強者解答
λt [ x1 ] [ v1 ] λt
令 X = V e , 更明白點 , [ x2 ] = [ v2 ] e
[ x3 ] [ v3 ]
代入 O.D.E.
λt λt
λ V e = A V e
=> 移項
λt
(A - λI) V e = 0
=> 特徵值問題
λ = 2 , 2 ,4
=> 特徵向量
λ1=2
[ 0 1 3 ][v1] [0] [1]
[ 0 0 1 ][v2] = [0] , V1 = k [0]
[ 0 0 2 ][v3] [0] [0]
特徵向量不夠用
2
(A-λI) V = 0
[ 0 0 7 ][v1] [0] [0]
[ 0 0 2 ][v2] = [0] , V2 = a [0] ← 解錯的XD
[ 0 0 4 ][v3] [0] [1]
眼睛快瞎了@@
[0] [1]
解出來應為 V2 = a [1] + a' [0]
[0] [0]
挑喜歡的數字當特徵向量吧!
λ3 = 4
[-2 1 3 ][v1] [0] [7]
[ 0 -2 1 ][v2] = [0] , V3 = b [2]
[ 0 0 0 ][v3] [0] [4]
(λ1)t (λ1)t (λ3) t
所以 X = c1 V1 e + c2 ( V2 + t V1 ) e + c3 V3 e
[1] 2t [0] [1] 2t [7] 4t
X(t) = c1 [0] e + c2 { [1] + [0] t } e + c3 [2] e
[0] [0] [0] [4]
At
= e C , C 是常數向量
3 ×1
2t 2t 4t
[ e te 7e ][c1]
= [ 2t 4t ][ ]
[ 0 e 2e ][c2]
[ 2t 4t ][ ]
[ 0 e 4e ][c3]
好久沒解了,有錯請指正,謝謝
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推 kilikolo0218:謝謝你 你幫我解了很多題過 真感謝阿 08/12 00:21
→ kilikolo0218:只不過我廣義特徵向量是用第一個向量去衍伸 08/12 00:30
→ kilikolo0218:出來是[0,1,0]T 08/12 00:30
→ kilikolo0218:也就是(A-λI) V=V1 補習班是這樣教的 但答案不一樣 08/12 00:36
→ kilikolo0218:想請教這答案是唯一解嗎??? 08/12 00:36
→ ntust661 :one thing is true... my answer is wrong XD 08/12 11:49
※ 編輯: ntust661 來自: 122.121.207.197 (08/12 11:52)