link this set of equations to a single unknown

1 回表示 (過去 30 日間)
semih beyçimen
semih beyçimen 2019 年 2 月 21 日
コメント済み: Rena Berman 2019 年 3 月 5 日
A=[a1 a2 a3 a4 0 0 0 0;a5 a6 a7 a8 0 0 0 0;0 0 0 0 a9 a10 a11 a12;0 0 0 0 a13 a14 a15 a16;a17 a18 a19 a20 a21 a22 a23 a24;a25 a26 a27 a28 a29 a30 a31 a32;a33 a34 a35 a36 a37 a38 a39 a40;a41 a42 a43 a44 a45 a46 a47 a48]
B=[C1;C2;C3;C4;C5;C6;C7;C8]
C=[0;0;0;0;0;0;0;0]
A*B=C
1- I want to take this determinant of A matrix and find its roots.
2- I wan to link this set of equations to a single unknown. ( like C1)
I'd be happy if you could help.
  1 件のコメント
Rena Berman
Rena Berman 2019 年 3 月 5 日
(Answers Dev) Restored edit

サインインしてコメントする。

回答 (1 件)

John D'Errico
John D'Errico 2019 年 2 月 21 日
編集済み: John D'Errico 2019 年 2 月 21 日
Not gonna happen, well, not easily. For multiple reasons. You have a totally symbolic matrix A? Or are those all fixed numbers? Even so, there are still issues.
If the coefficientts in A are symbolic unknowns, then computing the symbolic determinant will be computational nightmare. And then solving for a zero will be analytically impossible in mathematics, due to the Abel-Ruffini theorem.
If the coefficients in A are real numbers, then using the determinant is just a bad idea, taught to you by a teacher who did not understand why it was bad, who was in turn taught by other teachers. It is a bad meme that will live on forever.
So if A is a real matrix, then use null. The solution is given by the function null, if any solution exists.
If you want that in turn to be parameterized by C1 in a symbolic form, then you want to learn about the mathematical solution, and what the output of null means. That in turn can turn into a small course on linear algebra, depending on what you know.

カテゴリ

Help Center および File ExchangeNumeric Solvers についてさらに検索

製品


リリース

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by