I have a 4×4 matrix T, and I know its last row to be [0 0 0 1]. I also know these equations:
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];Tp_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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