線性代數考古題請高手幫忙

晚學想考數學研究所,請不吝指教了!

圖片參考:http://imgcld.yimg.com/8/n/AD01848228/o/2014012622...

Update:

非常謝謝您的指導。

這一題有人提供另一種算法,我不確定是否合理,可否請您指教?

他將矩陣A進行row reduction

得到結果是A=I

然後代入算式中,A^5-8A^4+19A^3-2A^2+18A-6I^3=22*I

請問這個答案對嗎?謝謝您了!Leon

4 Answers

Rating
  • 7 years ago
    Favorite Answer

    Consider the characteristic polynomial

    |A-pI|

    = (2-p)^2*(3-p)+2+2- 2(2-p) - 2(2-p) - (3-p)

    = -p^3+7p^2-11p+5

    So -A^3+7A^2-11A+5I = 0

    Also, consider another polynomial (p^5-8p^4+19a^3-2a^2+18a-6)

    by long division(or other possible means), we obtain

    (p^5-8p^4+19a^3-2a^2+18a-6)

    = (-p^2+p-1)(-p^3+7p^2-11p+5) + 7p^2+3p-1

    So

    A^5-8A^4+19A^3-2A^2+18A-6I

    = (-A^2+A-I)(-A^3+7A^2-11A+5I) + 7A^2+3A-I

    = 7A^2+3A - I

    (The rest is left to you since multiplication of matrices is quite easy, and I don't want to type out those matrices, most importantly xD)

    2014-01-26 23:28:10 補充:

    the first part is the application of Cayley-Hamilton theorem.

    2014-01-26 23:39:18 補充:

    sorry for my wrong calculations during long division, it should be

    (p^5-8p^4+19p^3-2p^2+18p-6)

    = (-p^2+p-1)(-p^3+7p^2-11p+5) + 21p^2+2p-1

    So

    A^5-8A^4+19A^3-2A^2+18A-6I

    = (-A^2+A-I)(-A^3+7A^2-11A+5I) + 21A^2+2A-I

    = 21A^2+2A - I

  • 萬分感謝!

    數學跟科學是相當精準的知識,這一點很吸引我。希望能夠跨入這些領域的大門,一窺自然的奧秘。

  • 7 years ago

    人家已經說了「想考數學研究所」

    ^__^

    2014-01-27 12:14:51 補充:

    我也祝你成功

    加油!加油!加油!

    ╭∧---∧╮

    │ .✪‿✪ │

    ╰/) ⋈ (\╯

  • 7 years ago

    好奇想問一下,你擅長領域是心理學,那大學的專業應該是心理學吧?

    心理學有牽涉到線性代數?

    2014-01-26 23:23:55 補充:

    原來如此。

    祝你成功

    加油!加油!加油!

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