Re: [線代] 線代(15)
※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: 試求
: [ 1 0 0 0 0 ... 0 ]
: [ -2 1 0 0 0 ... 0 ]
: [ 0 -2 1 0 0 ... 0 ]
: [ 0 0 -2 1 0 ... 0 ]
: [ . . . . . ... . ]
: [ 0 ... 0 0 -2 1 0 ]
: [ 0 ... 0 0 0 -2 1 ]
: 的反矩陣.
Ans:
Let
[ 1 0 0 0 ... 0 0 0 ]
[ -2 1 0 0 ... 0 0 0 ]
[ 0 -2 1 0 ... 0 0 0 ]
[ 0 0 -2 1 ... 0 0 0 ] = A
[ . . . . ... . . . ]
[ 0 0 0 0 ... -2 1 0 ]
[ 0 0 0 0 ... 0 -2 1 ]
If A is a matrix with order n
Then
[ 1 0 0 ... 0 ]
[ -2 1 0 ... 0 ]
[ (-2)^2 -2 1 ... 0 ]
A^(-1)=[ (-2)^3 (-2)^2 -2 ... 0 ]
[ . . . . ]
[ (-2)^(n-1) (-2)^(n-2) (-2)^(n-3) ... 1 ]
#
(Hint: 利用歸綜法)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.128.28.250
討論串 (同標題文章)