Trying to determine roots of this polynomial (Det)

k = [1750 -750 0; -750 1250 -500; 0 -500 500];

m = [75 0 0; 0 75 0; 0 0 50];
D=inv(m)*k;
syms x
I = [x,0,0;0,x,0;0,0,x];
Lambda = D-I;
DET = det(Lambda);
eqn = DET==0;
sol = root(eqn,3)

Answer 1

[John D'Errico]

k = [1750 -750 0; -750 1250 -500; 0 -500 500];
m = [75 0 0; 0 75 0; 0 0 50];
D=inv(m)*k;
syms x
I = [x,0,0;0,x,0;0,0,x];
Lambda = D-I;
DET = det(Lambda);
eqn = DET==0

eqn = 

At this point, you have generated a cubic polynomial. It is a symbolic polynomial in x. That would be a good start. You needed to make only one more step.

But, what do you think root does? (Nothing. There is no function named root.) You can use solve.

xsol = solve(eqn,'maxdegree',3)

xsol = 

And that looks pretty messy, but the fact is, the roots of a cubic polynomial are a bit messy for a completely general polynomial.

vpa(xsol)

ans = 

So there are three real roots. They look like they are complex roots, because they have an imaginary part, but it is an infinitessimal one. Just discard that part.

real(vpa(xsol))

ans = 

Those are the three roots. Are they correct?

eig(D)

ans = 3x1

31.6965 15.6084 2.6951

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Indeed, what you did was valid. At least until that very last line.

[Joseph Slattery]

ty

Answer 2

[Torsten]

k = [1750 -750 0; -750 1250 -500; 0 -500 500];
m = [75 0 0; 0 75 0; 0 0 50];
eig(k,m)

ans = 3x1

2.6951 15.6084 31.6965

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

[Joseph Slattery]

nice

Answer 3

[Sam Chak]

k   = [1750 -750 0; -750 1250 -500; 0 -500 500];
m   = [75 0 0; 0 75 0; 0 0 50];
D   = inv(m)*k;
syms x
I   = [x,0,0;0,x,0;0,0,x];
Lambda = D - I;
DET = det(Lambda);
eqn = DET==0;

Hi @Joseph Slattery

If you would like to use 'eqn' as the input argument for the 'roots()' command to find the roots of the polynomial, you can consider using this syntax:

sol = roots(fliplr(coeffs(lhs(eqn))))

sol = 

Essentially, it is equivalent to executing these four lines of code:

lhsEqn = lhs(eqn)           % get the left side of the symbolic equation

lhsEqn = 

C      = coeffs(lhsEqn)     % get the coefficients of the polynomial expression

C = 

P      = fliplr(C)          % flip the row vector C with the order of its elements reversed

P = 

sol    = roots(P)           % find the roots based on the coefficients in lhsEqn

sol =