Substitute symbolic matrices into numerical matrices

1 次查看（过去 30 天）
Ali Daher
Ali Daher 2022-9-25
评论： Torsten 2022-9-25
Let's say I am solving a linear system of Equations of the form: Ax=b at each timestep. Matrix A always always has the same structure, for instance A = [a b c; -a 2b c-d; -d 2c a]. During each timestep iteration I update the values of a, b, c ,d, but the structure of A (and of b) remains the same in terms of those symbols. Is there a way to construct a symbolic matrix and then just substitute the symbols with their updated values at each timestep rather than reconstruct the whole A matrix (and b vector) at each timestep?
Many thanks!

请先登录，再回答此问题。

回答（1 个）

Torsten
Torsten 2022-9-25
编辑：Torsten 2022-9-25
Ran in:
Do you know Cramer's rule ?
A = sym('A',[3 3]);
b = sym('b',[3 1]);
x = sym('x',[3 1]);
sol = solve(A*x==b,x)
sol = struct with fields:
x1: (A1_2*A2_3*b3 - A1_3*A2_2*b3 - A1_2*A3_3*b2 + A1_3*A3_2*b2 + A2_2*A3_3*b1 - A2_3*A3_2*b1)/(A1_1*A2_2*A3_3 - A1_1*A2_3*A3_2 - A1_2*A2_1*A3_3 + A1_2*A2_3*A3_1 + A1_3*A2_1*A3_2 - A1_3*A2_2*A3_1) x2: -(A1_1*A2_3*b3 - A1_3*A2_1*b3 - A1_1*A3_3*b2 + A1_3*A3_1*b2 + A2_1*A3_3*b1 - A2_3*A3_1*b1)/(A1_1*A2_2*A3_3 - A1_1*A2_3*A3_2 - A1_2*A2_1*A3_3 + A1_2*A2_3*A3_1 + A1_3*A2_1*A3_2 - A1_3*A2_2*A3_1) x3: (A1_1*A2_2*b3 - A1_2*A2_1*b3 - A1_1*A3_2*b2 + A1_2*A3_1*b2 + A2_1*A3_2*b1 - A2_2*A3_1*b1)/(A1_1*A2_2*A3_3 - A1_1*A2_3*A3_2 - A1_2*A2_1*A3_3 + A1_2*A2_3*A3_1 + A1_3*A2_1*A3_2 - A1_3*A2_2*A3_1)
And now please don't tell me that in reality, your matrix A is 50x50 :-)
  2 个评论
Torsten
Torsten 2022-9-25
Ran in:
And for such a matrix size, you use symbolic maths ? You will have much time to drink your coffee ...
Use numerical matrices and backslash without any structural analysis of the matrix:
A = rand(100);
b = rand(100,1);
x = A\b
x = 100×1
0.2890 0.3542 0.0288 -0.6519 -0.9014 1.1733 0.4072 0.8347 -0.2271 0.7861

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Linear Algebra 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by