# Section 9.8# Example 1 interface(imaginaryunit=i);

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with(linalg):with(Student[LinearAlgebra]):I := Id(3);

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

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evalm(A-r*I);

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det(%);

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factor(%);

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evalm(A-I);

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evalm((A-I)^2);

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evalm(exp(t)*(I + (A-I)*t));

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# Example 2A := matrix([[1,0,0],[1,3,0],[0,1,1]]);

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evalm(A-r*I);

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det(%);

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r1 := 1;

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u := vector(3);

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(A-r1*I)&*u;

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eq := evalm(%);

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eq1 := eq[1] = 0; eq2 := eq[2] = 0;eq3 := eq[3] = 0;

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#By inspection, we obtain the solutionu[1] := 0; u[2] := 0; u[3] := s;

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evalm(u);

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u1 := subs(s=1,evalm(u));

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x1 := evalm(exp(r1*t)*u1);

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u := vector(3);

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(A-r1*I)^2&*u;

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eq := evalm(%);

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eq1 := eq[1] = 0; eq2 := eq[2] = 0;eq3 := eq[3] = 0;

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u[2] := s;

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u[1] := solve(eq3,u[1]);

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u[3] := v;

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evalm(u);

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u2 := evalm(subs({s=1,v=0},evalm(u)));

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A-r1*I;

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evalm(%);

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evalm(% &* u2);

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evalm(u2 + %*t);

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x2 := evalm(exp(r1*t)*%);

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u := vector(3);

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r3 := 3;

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(A - r3*I)&*u;

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eq := evalm(%);

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eq1 := eq[1] = 0; eq2 := eq[2] = 0;eq3 := eq[3] = 0;

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# We readily obtain the solutionu[1] := 0;

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u[2] := 2*s;

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u[3] := s;

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evalm(u);

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u3 := subs(s=1,%);

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x3 := evalm(exp(r3*t)*u3);

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?augmentX := augment(x1,x2,x3);

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subs(t=0,evalm(X));

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X0 := simplify(%);

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X0inv := inverse(X0);

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evalm(X)&*evalm(X0inv);

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evalm(%);

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