フィルターのクリア

Solve matrix equations using loop

2 ビュー (過去 30 日間)
Alina Abdikadyr
Alina Abdikadyr 2022 年 10 月 8 日
コメント済み: Torsten 2022 年 10 月 8 日
Hello everyone!
I have a matrix A(3x3)
A= (2 10 15
3 5 -7
3 -2 -2 )
And Matrx B (3x1)
B = (B1
B2
B3)
Matrix C (3x1), where all values are equal to each other
C= (C1
C1
C1)
So, I need using loop, solve the equations and find the values of C1, B2 and B3. B1 is known for me, and has 125 elements. For each B1, I have to find C1, B2 and B3 respectively
Solve the equation A*B=C
Thank you in advance!
  3 件のコメント
Alina Abdikadyr
Alina Abdikadyr 2022 年 10 月 8 日
A*B=C

サインインしてコメントする。

回答 (1 件)

Chunru
Chunru 2022 年 10 月 8 日
The equations you have:
You want to solve it for , and . You need to rearrage the equations:
Now you solve this new system of equations with unknowns as :
A = [ 2 10 15
3 5 -7
3 -2 -2 ];
Anew = [-ones(3, 1) A(:, 2:3)];
AnewInv = inv(Anew);
b1 = rand(5,1); % rand(125,1) or your data
for i=1:length(b1)
bnew = -b1(i)*A(:, 1);
res = AnewInv*bnew %[c1; n2; b3]
end
res = 3×1
0.9814 0.0096 0.0134
res = 3×1
1.8311 0.0178 0.0250
res = 3×1
0.9554 0.0093 0.0130
res = 3×1
1.8897 0.0184 0.0258
res = 3×1
0.4087 0.0040 0.0056

カテゴリ

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

製品


リリース

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by