Find Eigenvalues and Eigenvectors of Matrix A

To find the eigenvalues, we need to solve the characteristic equation:

det(A - λI) = 0 where I is the identity matrix and λ is the eigenvalue.

A - λI = (2-λ) -1 2 5 (-3-λ) 3 -1 0 (−2−λ)

Taking the determinant, we get:

(2-λ)(-3-λ)(-2-λ) + 2(-1)(0) + (-1)(5)(-2-λ) - 2(-1)(-1)(-2-λ) - (2-λ)(0)(-1) Simplifying, we get: -λ^3 - 3λ^2 + 4λ - 2 = 0

This can be factored as: -(λ-2)(λ-1)(λ+1) = 0

Therefore, the eigenvalues are λ1 = 2, λ2 = 1, and λ3 = -1.

To find the eigenvectors, we need to solve the equation:

(A - λI)x = 0

For λ1 = 2:

(A - 2I)x = 0 -1 2 x1 0 5 -5 3 x * = 0 -1 0 -4 x3 0

We can simplify this to:

-x2 + 2x3 = 0 5x1 - 5x2 + 3x3 = 0 -x1 - 4x3 = 0

We can choose x3 = 1. Then, from the first and third equations, we get x1 = -4 and x2 = -2. Therefore, an eigenvector for λ1 = 2 is:

v1 = [-4, -2, 1]

For λ2 = 1:

(A - I)x = 1 -1 2 x1 0 5 -4 3 x * = 0 -1 0 -3 x3 0

We can simplify this to:

-x1 + x2 + 2x3 = 0 5x1 - 4x2 + 3x3 = 0 -x1 -3x3 = 0

We can choose x3 = 1. Then, from the first and third equations, we get x1 = -3 and x2 = -1. Therefore, an eigenvector for λ2 = 1 is:

v2 = [-3, -1, 1]

For λ3 = -1:

(A + I)x = 3 -1 2 x1 0 5 -2 3 x * = 0 -1 0 -1 x3 0

We can simplify this to:

3x1 - x2 + 2x3 = 0 5x1 - 2x2 + 3x3 = 0 -x1 - x3 = 0

We can choose x3 = 1. Then, from the first and third equations, we get x1 = 1 and x2 = 1. Therefore, an eigenvector for λ3 = -1 is:

v3 = [1, 1, 1]

Therefore, the eigenvalues and eigenvectors of A are:

Eigenvalues: λ1 = 2, λ2 = 1, λ3 = -1 Eigenvectors: v1 = [-4, -2, 1], v2 = [-3, -1, 1], v3 = [1, 1, 1]