Pure Mathematics - Matrix

2010-02-27 8:48 am
1. Find all the matrices X which satisfy the equation
X2 - 4X + 3I = 0, I being the unit matrix and 0 the null matrix.

2. If A and B are square matrices satisfying the equation A^2 = B, show that A and B commute.
Hence find matrices A such that A^2 = [1 1 0; 0 1 0; 2 0 1].

回答 (4)

2010-03-04 1:26 am
✔ 最佳答案
Pure Mathematics會出這些題目?因為我看來這些似乎是要使用台灣的工程數學的知識多於只需使用Pure Mathematics的知識才可解的題目。

2010-02-28 13:41:12 補充:
首先,對於第1題,我唔知道Cayley–Hamilton theorem(http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem)是否可逆。如果可逆還好一點,可以使用它;但如果不可逆,這題就注定要from first principle,即是設X = [a b; c d],然後就要解相關的聯立四元二次方程
 ╭
 │a² + bc - 4a + 3 = 0
_│ab + bd - 4b = 0
 │ac + cd - 4c = 0
 │bc + d² - 4d + 3 = 0
 ╰

2010-02-28 13:41:37 補充:
但是使用Cayley–Hamilton theorem的逆定理(如有的話)或是解聯立四元二次方程等舉動就已經不是Pure Mathematics期望要做的事。

2010-02-28 18:59:17 補充:
是哪一年或是哪裡出的題目咁大整蠱呀,要我們解非線性聯立方程(Pure Mathematics從不會教如何解非線性聯立方程)?

其實Pure Mathematics的題目是否容許出現任何關於非線性聯立方程的步驟?

2010-02-28 19:08:23 補充:
但對於第1題的那條聯立四元二次方程,並不是難解得這樣可憐,因為其中第2式和第3式都可以分解成AB = 0這樣式,而接著下來的步驟當然是(A = 0 and B = 0) or (A = k and B = 0) or (A = 0 and B = k) , where k is any non-zero complex number

這樣的步驟,稍為聰明的人都可以想到出來。

2010-03-01 13:15:24 補充:
To 1997年7月1日:
第1題,你的想法是否與我一樣都是from first principle,即是設X = [a b; c d],然後解相關的聯立四元二次方程
 ╭
 │a² + bc - 4a + 3 = 0
_│ab + bd - 4b = 0
 │ac + cd - 4c = 0
 │bc + d² - 4d + 3 = 0
 ╰


注意:答案未必需要與modal answer的完全一樣才算正確,只要計算過程正確,任意常數的數量足夠,驗算也驗出正確的話,就已經OK。

2010-03-01 13:35:57 補充:
至於第2題,我與你一樣亦有同感,都是get唔到「證明AB = BA」對「解matrix equation A² = B」有甚麼幫助。事實上,如果是可行的,類似的http://tw.knowledge.yahoo.com/question/question?qid=1010020803771亦不會直到現在還未得出結論。我亦不相信「證明AB = BA」對「解matrix equation A² = B」有任何幫助。

如果真係諗唔到的話,就當冇「Hence」這個字眼吧!

2010-03-01 13:54:59 補充:
對於第2題,就算找± [1 1 0; 0 1 0; 2 0 1]^(1/2),都不能保證能找出A² = [1 1 0; 0 1 0; 2 0 1]的所有解,而且找± [1 1 0; 0 1 0; 2 0 1]^(1/2)也必定需要牽涉把[1 1 0; 0 1 0; 2 0 1]進行diagonalization或是進行Jordan decomposition,更高超者可以使用Cayley–Hamilton theorem(http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem)及配合一些技巧代替前面的兩個方法,

2010-03-01 13:55:09 補充:
即是也必定需要牽涉找[1 1 0; 0 1 0; 2 0 1]的eigenvalues和eigenvectors(如需進行diagonalization或是進行Jordan decomposition)。

但這還是Pure Mathematics嗎?

2010-03-01 14:08:09 補充:
出現這樣的情況,唯有from first principle,即是設A = [p q r; s t u; v w x],然後就要解相關的聯立九元二次方程
 ╭
 │p² + qs + rv = 1
 │pq + qt + rw = 1
 │pr + qu + rx = 0
 │ps + st + uv = 0
─┤qs + t² + uw = 1
 │rs + tu + ux = 0
 │pv + sw + vx = 2
 │qv + tw + wx = 0
 │rv + uw + x² = 1
 ╰

2010-03-02 11:31:01 補充:
To: 子路
003號意見的那個方法就已經是不需要使用Cayley–Hamilton theorem的方法(笑!)

我亦看過Pure Mathematics書,發覺它亦有一些題目是需要解聯立多元高次方程。由此看來,我諗Pure Mathematics不容許的只是分析聯立多元高次方程其解的性質而不是純粹解聯立多元高次方程。

2010-03-03 17:26:31 補充:

圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/matrixequation/matrixequation00.jpg


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圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/matrixequation/matrixequation19.jpg

參考資料:
my wisdom of pure maths
2010-03-14 10:09 pm
hjyytytuytuyu
2010-03-02 2:59 am
for 1
is X restricted to be a 2x2 one?

2010-03-01 18:25:30 補充:
我想到第2點做
第1點都淨係諗到2個soln....
問題在於, PM不需要學cayleyhamilton的
注意,不能用cayleyhamilton去找I和3I的~

2010-03-01 18:59:47 補充:
for question one
i am thinking how to avoid using Cayleyhamilton thm
(anyway this question should not appear in AL or you are guided to do it)

for question two, see:



If A^2=B
AB=A(A^2)=A^3
BA=(A^2)A=A^3

so A,B commute

圖片參考:http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-03-01-18-56-53.png


2010-03-01 19:00:22 補充:
a,b,h are any numbers in the complex set

2010-03-02 12:09:17 補充:
ha
i missed the key step
squre the A out
you get a^2=1 ab=0.5 b^2=-ha
solve it yourself~

2010-03-02 12:09:37 補充:
see my solution for question 1
the reason why i don't want to tackle it by
"first principle" is that i don't want to solve those eqts....
so I compare the inverse of them!!

http://i256.photobucket.com/albums/hh182/zilu_photo/img035.jpg

2010-03-02 12:09:39 補充:
By the way i really want to know what is your teacher's suggest method!!
2010-03-01 12:24 am
謝謝 doraemonpaul 的意見!
題目真的是 pure mathematics 的題目.
建議答案是[3, 0; 0, 3] , [1, 0; 0, 1], [3, 0; 0, 1], [1, 0; 0, 3],
[2+p, q(1+/-p); (1+/-p)/q, 2-p] where p, q are real and q is not 0.

2010-03-01 03:45:00 補充:
第1題, 我得出前4個答案, 但最後一個找不到...
第2題, 第2部分應如何開始?

2010-03-01 20:17:29 補充:
首先, 真的要多謝以上兩位的幫忙!
真的感到非常高興!
To "doraemonpaul",
我第一題的想法和你的一樣!


收錄日期: 2021-04-13 17:06:54
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