Solving for unknown matrix?
46 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I have three known matrices, and one unknown matrix that I need to solve for. The matrices are defined in the code below. The equation to find phi is G = C*phi*B. How do I solve for phi? I tried doing phi = inv(C)*G*inv(B) but C and B are not square matrices.
syms t
G = [10*exp(-t) 3*exp(-t); 0 exp(-2*t)*cos(3*t)+2/3*exp(-2*t)*sin(3*t)];
C = [13 4 1; 1 1 0];
B = [1 0; -1 1; 1 -1];
0 Kommentare
Antworten (3)
David Goodmanson
am 26 Okt. 2017
Hi Andrew,
I understand that you are looking for a symbolic solution, but if G were a matrix of numbers, then
phi = C\G/B
is the fastest way to find a phi that works. For your matrix dimensions, phi always has four nonzero elements, but they are not necessarily in the upper left corner of the 3x3.
As Walter has mentioned, the solution for psi is far from unique. For more solutions,
nullC = null(C)
nullBtt = null(B.').'
c1 = rand(1,3); % c1 is 1x3 vector of arbitrary constants; rand here for demo
c2 = rand(3,1); % c2 is 3x1 vector of arbitrary constants; rand here for demo
arb1 = nullC*c1;
arb2 = c2*nullBtt;
phi_new = phi + arb1 + arb2 % another solution
0 Kommentare
Walter Roberson
am 16 Okt. 2017
phi = sym('phi', [3 3]);
one_solution = solve(G == C*phi*B, phi(1:2,1:2));
Now one_solution.phi1_1, .phi1_2, .phi2_1, and .phi2_2 give definitions for the top left corner of phi in terms of the other entries of phi . That is, the system is underdetermined and there are an infinite number of solutions.
0 Kommentare
Zahid Ullah
am 17 Dez. 2022
2 Kommentare
DGM
am 17 Dez. 2022
% Ax + B = R
A = [5 -2; 3 -2];
B = [3 5; 4 -1];
R = [7 0; 4 -8];
% solve for x
x = A\(R-B)
% back-calculate R, check
R2 = A*x + B
John D'Errico
am 17 Dez. 2022
@Zahid Ullah please don't post questions as an anser to a completely different question. Learn to ask a question, not post an answer.
Siehe auch
Kategorien
Mehr zu Mathematics and Optimization finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!