# Section 9.6# Example 1 interface(imaginaryunit=i);

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with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

with(Student[LinearAlgebra]):I := Id(2);

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A := matrix([[-1,2],[-1,-3]]);

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evalm(A-r*I);

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det(%);

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solve(%=0,r);

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r1 := %[1];

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z := vector(2);

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eq := evalm((A-r1*I)&*z);

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eq1 := eq[1]=0; eq2 := eq[2]=0;

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# Although they don't look like it, these equations are actually multiples of each other.#If we apply Gaussian elmination, we geteq2 := eq2 + eq1/(1-i);

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eq2 := simplify(%);

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eq1;eq2;

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# This is easily solved by letting z[1] = 2s and solving for z[2]:z[1] := 2*s;

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z[2] := solve(eq1,z[2]);

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evalm(z);

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subs(s=1,%);

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z1 := evalm(%); #Eigenvector corresponding to r1

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alpha := Re(r1); beta := Im(r1); a := evalm(Re(z1)); b := evalm(Im(z1));

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x := exp(alpha*t)(c1*(cos(beta*t)*a-sin(beta*t)*b) + c2*(sin(beta*t)*a+cos(beta*t)*b));

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evalm(%);

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# Finding the eigen-pairs another way:eigenvects(A);

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#Using dsolve to find the general solution:de1 := diff(x1(t),t) = -x1(t)+2*x2(t);

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de2 := diff(x2(t),t) = -x1(t)-3*x2(t);

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dsolve({de1,de2});

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# Notice that if _C1 is changed to 2*c2 and _C2 is changed to 2*c1, dsolve's answer is# the same as the first answer above.# Example 2 A := matrix([[0,1,0,0],[-(k1+k2)/m1,0,k2/m1,0],[0,0,0,1],[k2/m2,0,-(k2+k3)/m2,0]]);

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evalm(subs({m1=1,m2=1,k1=1,k2=2,k3=3},evalm(A)));

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A_Ex2 := (%);

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I := Id(4);

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evalm(A_Ex2 - r*I);

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det(%);

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subs(r=sqrt(z),%);

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rsquared := solve(%,z);

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beta1 := sqrt(-rsquared[1]);beta2 := sqrt(-rsquared[2]);

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f1 := evalf(beta1/(2*Pi));

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f2 := evalf(beta2/(2*Pi));

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