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To find the eigenvalues and eigenvectors, we need to solve the characteristic equation:
|A - λI| = 0
where A is the given matrix, I is the identity matrix, λ is the eigenvalue, and |.| denotes the determinant.
So, we have:
|1-λ 0 1| |0 2-λ 3| |0 0 3-λ| = 0
Expanding the determinant along the first row, we get:
(1-λ) [(2-λ)(3-λ) - 0(0)] - 0[0(3-λ) - 0(0)] + 1[0(0) - 0(2-λ)] = 0
Simplifying the above equation, we get:
(1-λ)(λ^2 - 5λ + 6) = 0
This quadratic equation has three roots: λ = 1, λ = 3, and λ = 2.
Now, we find the eigenvectors for each eigenvalue by solving the equation:
(A - λI)x = 0
where x is the eigenvector.
For λ = 1, we have:
(A - λI)x = (A - I)x = 0
Substituting λ = 1 and solving the equation, we get:
x1 = -x3
So, the eigenvector corresponding to λ = 1 is:
v1 = [1, 0, -1]
For λ = 3, we have:
(A - λI)x = (A - 3I)x = 0
Substituting λ = 3 and solving the equation, we get:
x2 + 3x3 = 0
So, the eigenvector corresponding to λ = 3 is:
v2 = [0, -3, 1]
For λ = 2, we have:
(A - λI)x = (A - 2I)x = 0
Substituting λ = 2 and solving the equation, we get:
x3 = 0
So, the eigenvector corresponding to λ = 2 is:
v3 = [1, -3/2, 0]
Therefore, the eigenvalues and eigenvectors of the given matrix are:
Eigenvalues: λ1 = 1, λ2 = 3, and λ3 = 2
Eigenvectors: v1 = [1, 0, -1], v2 = [0, -3, 1], and v3 = [1, -3/2, 0]
