Solve system of equations with the symbolic matrix

I have a 4x4 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];
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?

Answer 1

[Walter Roberson]

T = sym('x', [4 4]);
T(4,:) = [0 0 0 1];
T

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,[]))

s = struct with fields:
    x1_1: [1×1 sym]
    x2_1: [1×1 sym]
    x3_1: [1×1 sym]
    x1_2: [1×1 sym]
    x2_2: [1×1 sym]
    x3_2: [1×1 sym]
    x1_4: [1×1 sym]
    x2_4: [1×1 sym]
    x3_4: [1×1 sym]

subs(T,s)

ans = 

[Cengizhan Demirbas]

Thanks, but I also wonder how can I generalize this method? For example, if i have different pold and pnew equations that result in unsolved 2nd column, how can i get it symbolically without specifically solving for other colums?

[Walter Roberson]

T = sym('x', [4 4]);
T(4,:) = [0 0 0 1];
T

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]

M = 

s = solve(M)

s = struct with fields:
    x1_1: [1×1 sym]
    x1_2: [1×1 sym]
    x1_4: [1×1 sym]
    x2_1: [1×1 sym]
    x2_2: [1×1 sym]
    x2_4: [1×1 sym]
    x3_1: [1×1 sym]
    x3_2: [1×1 sym]
    x3_4: [1×1 sym]

subs(T,s)

ans = 

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)

A = 

b = 

vars(:) == A\b

ans = 

[Cengizhan Demirbas]

Thanks!!