How do I found A-B*x which these matrix are 3x3

1 回表示 (過去 30 日間)
Rommel Brito
Rommel Brito 2020 年 9 月 9 日
回答済み: Bruno Luong 2020 年 9 月 11 日
A=[1231200 -615600 0; -615600 1231200 -615600; 0 -615600 615600]
A =
1231200 -615600 0
-615600 1231200 -615600
0 -615600 615600
B=[62.68 0 0; 0 62.68 0; 0 0 62.68]
B =
62.6800 0 0
0 62.6800 0
0 0 62.6800
And I want to get the A-Bx which x is a uknown variable that I want to know.

回答 (2 件)

Walter Roberson
Walter Roberson 2020 年 9 月 9 日
>> x=B\A
x =
19642.6292278239 -9821.31461391193 0
-9821.31461391193 19642.6292278239 -9821.31461391193
0 -9821.31461391193 9821.31461391193
>> A-B*x
ans =
0 0 0
0 0 0
0 0 0
  3 件のコメント
Walter Roberson
Walter Roberson 2020 年 9 月 10 日
A = [1231200, -615600, 0; -615600, 1231200, -615600; 0, -615600, 615600];
B = [62.68 0 0; 0 62.68 0; 0 0 62.68];
syms L
vals = solve(det(A - B*L), 'MaxDegree', 3);
real(vpa(vals)) %imaginary component is noise
Rommel Brito
Rommel Brito 2020 年 9 月 10 日
Thank you! That worked!

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Bruno Luong
Bruno Luong 2020 年 9 月 11 日
Waoh, revise the theory of eigen vectors and linear algebra in general
Multiply a vector by a constant diagonal matrix is like multiply by the scalar.
This equation admits a solution for only when B = lambda*Identity where lambda is eigen value and x is the corresponding eigen eigen vector.
I'm surprises no one tell you that.

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