How do I know, what simplifying steps MATLAB had taken?
I know that using "simplify" command I can simplify my trignometric matrices. But, can we take output on each simplification step that MATLAB takes to simplify the given expression?
I am trying to simplify this:
if true
% clc
clear all
syms o l1 l2 l0;
A=[ cos(o) -sin(o) 1;
cos(o-(2*pi/3)) -sin(o-(2*pi/3)) 1;
cos(o-(4*pi/3)) -sin(o-(4*pi/3)) 1;]
invA= (2/3)*[cos(o) cos(o-2*pi/3) cos(o-4*pi/3);
-sin(o) -sin(o-2*pi/3) -sin(o-4*pi/3);
0.5 0.5 0.5]
Lss = [l0 + ((l2)*cos(2*o)) (-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) (-0.5*l1)+((l2)*cos((2*o)+2*pi/3));
(-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) l0 + ((l2)*cos(2*(o-2*pi/3))) (-0.5*l1)+((l2)*cos(2*o))
(-0.5*l1)+((l2)*cos((2*o)+2*pi/3)) (-0.5*l1)+((l2)*cos((2*o))) l0 + ((l2)*cos(2*(o-4*pi/3)))]
X=(invA)*(Lss)*(A)
W=simplify(X)
end
and the answer comes to be:
if true
% W =[ l0 + l1/2 + (3*l2)/2, 0, 0]
[ 0, l0 + l1/2 – (3*l2)/2, 0]
[ 0, 0, l0 – l1]
end
which is true. But how do MATLAB got there??I know that using "simplify" command I can simplify my trignometric matrices. But, can we take output on each simplification step that MATLAB takes to simplify the given expression?
I am trying to simplify this:
if true
% clc
clear all
syms o l1 l2 l0;
A=[ cos(o) -sin(o) 1;
cos(o-(2*pi/3)) -sin(o-(2*pi/3)) 1;
cos(o-(4*pi/3)) -sin(o-(4*pi/3)) 1;]
invA= (2/3)*[cos(o) cos(o-2*pi/3) cos(o-4*pi/3);
-sin(o) -sin(o-2*pi/3) -sin(o-4*pi/3);
0.5 0.5 0.5]
Lss = [l0 + ((l2)*cos(2*o)) (-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) (-0.5*l1)+((l2)*cos((2*o)+2*pi/3));
(-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) l0 + ((l2)*cos(2*(o-2*pi/3))) (-0.5*l1)+((l2)*cos(2*o))
(-0.5*l1)+((l2)*cos((2*o)+2*pi/3)) (-0.5*l1)+((l2)*cos((2*o))) l0 + ((l2)*cos(2*(o-4*pi/3)))]
X=(invA)*(Lss)*(A)
W=simplify(X)
end
and the answer comes to be:
if true
% W =[ l0 + l1/2 + (3*l2)/2, 0, 0]
[ 0, l0 + l1/2 – (3*l2)/2, 0]
[ 0, 0, l0 – l1]
end
which is true. But how do MATLAB got there?? I know that using "simplify" command I can simplify my trignometric matrices. But, can we take output on each simplification step that MATLAB takes to simplify the given expression?
I am trying to simplify this:
if true
% clc
clear all
syms o l1 l2 l0;
A=[ cos(o) -sin(o) 1;
cos(o-(2*pi/3)) -sin(o-(2*pi/3)) 1;
cos(o-(4*pi/3)) -sin(o-(4*pi/3)) 1;]
invA= (2/3)*[cos(o) cos(o-2*pi/3) cos(o-4*pi/3);
-sin(o) -sin(o-2*pi/3) -sin(o-4*pi/3);
0.5 0.5 0.5]
Lss = [l0 + ((l2)*cos(2*o)) (-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) (-0.5*l1)+((l2)*cos((2*o)+2*pi/3));
(-0.5*l1)+((l2)*cos((2*o)-2*pi/3)) l0 + ((l2)*cos(2*(o-2*pi/3))) (-0.5*l1)+((l2)*cos(2*o))
(-0.5*l1)+((l2)*cos((2*o)+2*pi/3)) (-0.5*l1)+((l2)*cos((2*o))) l0 + ((l2)*cos(2*(o-4*pi/3)))]
X=(invA)*(Lss)*(A)
W=simplify(X)
end
and the answer comes to be:
if true
% W =[ l0 + l1/2 + (3*l2)/2, 0, 0]
[ 0, l0 + l1/2 – (3*l2)/2, 0]
[ 0, 0, l0 – l1]
end
which is true. But how do MATLAB got there?? simplify, steps, taken, what, how MATLAB Answers — New Questions
