How to find solutions for a determinat with 4 variables equal to zero

I have a matrix 3x3 with 4 variables.I want to find for which values of these variables the determinant is equal to zero.

The matrix and code is here:

q1=sym('q1');
q2=sym('q2');
q3=sym('q3');
q4=sym('q4');

l1 = 250;
l2 = 220;
l3 = 160;
lE = 214.6;

Jac=[-sin(q1)*(l3*cos(q2+q3)+l2*cos(q2)+lE*sin(q2+q3+q4)) -cos(q1)*(l2*sin(q2)+l3*sin(q2+q3)-lE*cos(q2+q3+q4)) -cos(q1)*(l3*sin(q2+q3)-lE*cos(q2+q3+q4));cos(q1)*(l3*cos(q2+q3)+l2*cos(q2)+lE*sin(q2+q3+q4)) -sin(q1)*(l3*sin(q2+q3)+l2*sin(q2)+lE*cos(q2+q3+q4)) -sin(q1)*(l3*sin(q2+q3)+lE*cos(q2+q3+q4));0 l3*cos(q2+q3)+l2*cos(q2)+lE*sin(q2+q3+q4) l3*cos(q2+q3)+lE*sin(q2+q3+q4)]

I know that the solution may be q3=0 ^ q4=pi/2.

If i do

solve(Jac==0)

i have not the desirable solution.

[John D'Errico]

Please don't make repeated posts of the same question.

[Walter Roberson]

This is not the same as http://www.mathworks.com/matlabcentral/answers/283584-how-to-find-the-values-of-variables-that-a-matrice-6x5-looses-rank -- this one involves the determinant of a 3 x 3 matrix and the other one involves a 6 x 5 matrix. They are, though, related.

[John D'Errico]

Virtually identical. The constants are the same, as well as many of the terms in the equations themselves. Knowing how to find the solution to one would show how to solve the other.

[Pyrros Tsopanakis]

I am sorry if my question is wrong for the forum but i am new here. I posted as different question because somebody told me that matlab cannot solve a matrix with so many (6x5 and 4variables) variables. So i posted as different in order to learn if it is possible to solve it with matlab.

Answer 1

[Walter Roberson]

"With difficulty" ?

The degenerate solutions include:

q2 = 0, q4 = -q3-arcsin(1100/1073+(800/1073)*cos(q3))

q3 = 0, q2 = -arctan((1/1073)*(1900+1073*sin(q4))/cos(q4))

q3 = 0, q4 = -q2-arcsin((1900/1073)*cos(q2))

q4 = 0, q2 = -arctan((1100+800*cos(q3)+1073*sin(q3)) / (-800*sin(q3)+1073*cos(q3)))

q4 = 0, q3 = arctan(-(1180300/1791329)*cos(q2)+(800/1791329)*sqrt(-1210000*cos(q2)^2+1791329), -(880000/1791329)*cos(q2)-(1073/1791329)*sqrt(-1210000*cos(q2)^2+1791329))-q2

q4 = 0, q3 = arctan(-(1180300/1791329)*cos(q2)-(800/1791329)*sqrt(-1210000*cos(q2)^2+1791329), -(880000/1791329)*cos(q2)+(1073/1791329)*sqrt(-1210000*cos(q2)^2+1791329))-q2

q2 = pi/2, q3 = arctan(1073*cos(q4)/(1073*sin(q4)+800))

q2 = pi/2, q4 = -q3+arccos((800/1073)*sin(q3))

q3 = pi/2, q2 = arctan((1100+1073*cos(q4))/(1073*sin(q4)+800))

q4 = pi/2, q2 = arctan((1/1873)*(1100+1873*cos(q3))/sin(q3))

q4 = pi/2, q3 = -q2+Pi-arccos((1100/1873)*cos(q2))

q1 = 0, q3 = 0, q4 = Pi/2

q2 = 0, q3 = 0, q4 = Pi/2

q2 = Pi, q3 = 0, q4 = Pi/2

... and possibly other degenerate solutions.

I say "degenerate" here in the sense that these are solutions regardless of the (finite) values of any variable not mentioned in the list: solutions that do not depend upon all 4 variables.

There are a number of other solutions involving all 4 variables. For example,

q1 = 0, q2 = -arctan((1073*sin(q3)*cos(q4)+1073*cos(q3)*sin(q4)+800*cos(q3)+1100) / (1073*cos(q3)*cos(q4)-1073*sin(q3)*sin(q4)-800*sin(q3)))

I did not do anything special here: I just went through a series of cases. Finding the complete set of solutions takes a bunch of computing time and more memory than I have on my system.