Find the Eigenvalues

Problem

det(1−λ)=0

Solution

1. Set up the characteristic equation by subtracting λ from the diagonal elements of the matrix A and setting the determinant of A−λ*I to zero.

2. Expand the determinant using cofactor expansion along the first row.

|A−λ*I|=(1−λ)*det(3−λ)+5*det(1)−3*det(1)

3. Calculate the first 3×3 determinant.

(1−λ)*[(3−λ)*(−λ*(1−λ)−1)−2*(1−λ−1)]

(1−λ)*[(3−λ)*(λ2−λ−1)+2*λ]

(1−λ)*[−λ3+4*λ2−2*λ−3+2*λ]=λ4−5*λ3+4*λ2+3*λ−3

4. Calculate the second 3×3 determinant multiplied by 5.

5*[1*(1−λ−1)−(3−λ)*(3*(1−λ)−0)]

5*[−λ−(3−λ)*(3−3*λ)]=5*[−λ−(9−12*λ+3*λ2)]

−15*λ2+55*λ−45

5. Calculate the third 3×3 determinant multiplied by −3

−3*[1*(1+λ)−(3−λ)*(3−0)+2*(3−0)]

−3*[1+λ−9+3*λ+6]=−3*[4*λ−2]=−12*λ+6

6. Combine all terms to find the characteristic polynomial.

P(λ)=λ4−5*λ3+4*λ2+3*λ−3−15*λ2+55*λ−45−12*λ+6

P(λ)=λ4−5*λ3−11*λ2+46*λ−42

7. Solve for the roots of the polynomial. By testing small integers or using numerical methods, we find λ=3 and λ=√(,7),−√(,7) do not work directly, but factoring reveals roots.

P(λ)=(λ−3)*(λ3−2*λ2−17*λ+14)

Testing λ=2 8−8−34+14≠0 Testing λ=−4.47,0.78,5.69 approximately.
The roots are (λ_1)=3 (λ_2)≈0.783 (λ_3)≈5.691 (λ_4)≈−4.474

Final Answer

λ={3,≈0.783,≈5.691,≈−4.474}

Want more problems? Check here!