Given:
The eigenvalues of matrix A are 1, -2, and 3.

To find:
The eigenvalues of the matrices 3I - 2A and A^2.

Solution:

1. Eigenvalues of 3I - 2A:
To find the eigenvalues of the matrix 3I - 2A, where I is the identity matrix, we need to subtract 2A from the scalar multiple of the identity matrix.

Let λ be the eigenvalue of 3I - 2A and v be the corresponding eigenvector.

(3I - 2A)v = λv

Rearranging the equation, we get:

(3I - 2A - λI)v = 0

Simplifying further, we have:

(3 - λ)I - 2A)v = 0

Since v is a non-zero vector (eigenvector), the determinant of (3 - λ)I - 2A must be zero.

Det((3 - λ)I - 2A) = 0

Expanding the determinant, we get:

(3 - λ)^2(3 - λ) - 2(3 - λ)(2A - λI) = 0

Simplifying further, we have:

(3 - λ)^3 - 2(3 - λ)(2A - λI) = 0

Expanding and simplifying, we get:

(3 - λ)^3 - 2(6 - 2λ)A + 2(3 - λ)λI = 0

Since A has eigenvalues 1, -2, and 3, we can substitute the eigenvalues into the equation and solve for λ.

Let's substitute the eigenvalue 1:

(3 - 1)^3 - 2(6 - 2(1))A + 2(3 - 1)(1)I = 0

Simplifying further, we get:

2^3 - 2(6 - 2)A + 2(3 - 1)(1)I = 0

8 - 2(4)A + 2(2)(1)I = 0

8 - 8A + 4I = 0

8A - 4I = 8

4(A - 2I) = 8

A - 2I = 2

Since A - 2I is singular, the eigenvalue 1 is not valid.

Let's substitute the eigenvalue -2:

(3 - (-2))^3 - 2(6 - 2(-2))A + 2(3 - (-2))(-2)I = 0

(3 + 2)^3 - 2(6 + 4)A + 2(3 + 2)(-2)I = 0

5^3 - 2(10)A + 2(5)(-2)I = 0

125 - 20A - 20I = 0

20A + 20I = 125

4(A + I) = 25

A + I = 6.25

Since A + I is singular, the eigenvalue -2 is not valid.

Let