Re: [線代] 兩題行列式(台大資工)

[Sfly (topos)]

※ 引述《mqazz1 (無法顯示)》之銘言：
: 1. suppose A in R^(3*3)
: and det(xI - A) = x^3 - x^2 + 3x - 2,
: then det(xI - A^2) = ?

det(t^2I - A^2)=det(tI-A)det(tI+A) = -det(tI-A)det((-t)I-A)
               = (t^3-t^2+3t-2)(t^3+t^2+3t+2)
               =(t^3+3t)^2 - (t^2+2)^2
               =t^6+2t^4+9t^2-4.

so, det(xI - A^2) = x^3+2x^2+9x-4.

: 2.  [2^0 2^1 2^2 2^3 2^4 2^5]
:     [2^1 2^0 2^1 2^2 2^3 2^4]
: T = [2^2 2^1 2^0 2^1 2^2 2^3]   //2^3 表 2的3次方=8
:     [2^3 2^2 2^1 2^0 2^1 2^2]
:     [2^4 2^3 2^2 2^1 2^0 2^1]
:     [2^5 2^4 2^3 2^2 2^1 2^0]
: det(T) = ?
: 請問這兩題怎麼解呢?
: 謝謝


add the i th row to the 6-i th row  for i=4,5,6
minus the j th cloumn from the 6-j th column for j=1,2,3.
then
        |1+2^5    2+2^4  2^2+2^3||1-2  2-4  4-8  |
det T = |2+2^4    1+2^3  2+2^2  ||2-4  1-8  2-16 | = 3^6.
        |2^2+2^3  2+2^2  1+2    ||4-8  2-16 1-32 |

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※ 編輯: Sfly            來自: 76.94.119.209        (02/23 11:48)
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