Find the Eigenvalues

Problem

det(3−λ)=0

Solution

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

det(A−λ*I)=0

2. Expand the determinant along the fourth column, which contains three zeros, to simplify the calculation.

(2−λ)*det(3−λ)=0

3. Calculate the 3×3 determinant using expansion along the first row.

(2−λ)*[(3−λ)*((1−λ)*(2−λ)+24)+2*(−15−3*(1−λ))]=0

4. Simplify the expression inside the brackets.

(2−λ)*[(3−λ)*(λ2−3*λ+26)+2*(3*λ−18)]=0

5. Distribute and combine like terms to find the characteristic polynomial.

(2−λ)*[−λ3+6*λ2−35*λ+78+6*λ−36]=0

(2−λ)*[−λ3+6*λ2−29*λ+42]=0

6. Factor the cubic polynomial by testing possible roots; λ=2 is a root since −8+24−58+42=0

(2−λ)*[−(λ−2)*(λ2−4*λ+21)]=0

(λ−2)2*(λ2−4*λ+21)=0

7. Solve for the remaining roots using the quadratic formula on λ2−4*λ+21=0

λ=(4±√(,16−84))/2

λ=(4±√(,−68))/2

λ=2±i√(,17)

Final Answer

λ=2,2,2+i√(,17),2−i√(,17)

Want more problems? Check here!