Find the Eigenvalues [[1,-1,-2],[1,3,2],[1,-1,2]]

Problem

[[1,−1,−2],[1,3,2],[1,−1,2]]

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.

det(A−λ*I)=0

2. Write the determinant expression for the matrix.

|[1−λ,−1,−2],[1,3−λ,2],[1,−1,2−λ]|=0

3. Expand the determinant along the first row.

(1−λ)*|[3−λ,2],[−1,2−λ]|−(−1)*|[1,2],[1,2−λ]|+(−2)*|[1,3−λ],[1,−1]|=0

4. Calculate the 2×2 determinants.

(1−λ)*((3−λ)*(2−λ)−(−2))+(1)*((2−λ)−2)−2*(−1−(3−λ))=0

5. Simplify the terms inside the parentheses.

(1−λ)*(λ2−5*λ+6+2)+(−λ)−2*(λ−4)=0

(1−λ)*(λ2−5*λ+8)−λ−2*λ+8=0

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

λ2−5*λ+8−λ3+5*λ2−8*λ−3*λ+8=0

−λ3+6*λ2−16*λ+16=0

λ3−6*λ2+16*λ−16=0

7. Factor the polynomial. By testing small integers, we find λ=2 is a root since 8 - 24 + 32 - 16 = 0$.

(λ−2)*(λ2−4*λ+8)=0

8. Solve for the remaining roots using the quadratic formula for λ2−4*λ+8=0

λ=(4±√(,(−4)2−4*(1)*(8)))/(2*(1))

λ=(4±√(,16−32))/2

λ=(4±√(,−16))/2

λ=2±2*i

Final Answer

λ=2,2+2*i,2−2*i

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