Solve system of equations with the symbolic matrix
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I have a 4x4 matrix T, and I know its last row to be [0 0 0 1]. I also know these equations:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/532959/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/532964/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/532969/image.png)
After solving these by hand, I find that third column has no solution (except for [4,3], which i know to be 0).
My question is how can I keep these unsolved variables symbolic while solving the system?
I tried the following code:
T = sym('x', [4 4]);
T(4,:) = [0 0 0 1];
T
p_new1 = [2 0 0 1].';
p_old1 = [0 0 0 1].';
p_new2 = [3 0 0 1].';
p_old2 = [1 0 0 1].';
p_new3 = [2 0 1 1].';
p_old3 = [0 1 0 1].';
eqn1 = p_new1 == T*p_old1;
eqn2 = p_new2 == T*p_old2;
eqn3 = p_new3 == T*p_old3;
sol = solve([eqn1, eqn2, eqn3])
This results in a strut with 9 elements that are solved. It completely excludes third column which i want to remain symbolic like x13 x23 x33. How can I do this?
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采纳的回答
Walter Roberson
2021-2-26
T = sym('x', [4 4]);
T(4,:) = [0 0 0 1];
T
p_new1 = [2 0 0 1].';
p_old1 = [0 0 0 1].';
p_new2 = [3 0 0 1].';
p_old2 = [1 0 0 1].';
p_new3 = [2 0 1 1].';
p_old3 = [0 1 0 1].';
eqn1 = p_new1 == T*p_old1;
eqn2 = p_new2 == T*p_old2;
eqn3 = p_new3 == T*p_old3;
s = solve([eqn1, eqn2, eqn3], reshape(T(1:3,[1:2,4]),1,[]))
subs(T,s)
3 个评论
Walter Roberson
2021-2-26
T = sym('x', [4 4]);
T(4,:) = [0 0 0 1];
T
p_new1 = [2 0 0 1].';
p_old1 = [0 0 0 1].';
p_new2 = [3 0 0 1].';
p_old2 = [1 0 0 1].';
p_new3 = [2 0 1 1].';
p_old3 = [0 1 0 1].';
eqn1 = p_new1 == T*p_old1;
eqn2 = p_new2 == T*p_old2;
eqn3 = p_new3 == T*p_old3;
M = [eqn1, eqn2, eqn3]
s = solve(M)
subs(T,s)
You can see here that I didn't have to do anything special -- solve() automatically worked it out in terms of variables actually present, and subs() back into T is enough to get nice matrix form.
But if you want more certainty:
vars = symvar(M);
[A,b] = equationsToMatrix(M, vars)
vars(:) == A\b
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