For each of the following matrices, and eigenvalue λ has been given.[ 0 4−1 5], λ = 4(a)[2 22 −1], λ = −2(b)3 1 −10 1 24 2 0, λ = 3(c)Find one eigenvector corresponding to the given eigenvalue
Question
For each of the following matrices, and eigenvalue λ has been given.[ 0 4−1 5], λ = 4(a)[2 22 −1], λ = −2(b)3 1 −10 1 24 2 0, λ = 3(c)Find one eigenvector corresponding to the given eigenvalue
Solution
Sure, let's find the eigenvectors for each of the given matrices and eigenvalues.
(a) For the matrix [0 4; -1 5] and λ = 4, we first subtract λ from the diagonal entries of the matrix to get the matrix [0-4 4; -1 5-4] = [-4 4; -1 1].
We then solve the system of linear equations given by this matrix, which is -4x + 4y = 0 and -x + y = 0. Both equations simplify to y = x. So, any vector where the x and y components are equal will be an eigenvector. For simplicity, we can choose x = 1, so y = 1. Therefore, an eigenvector is [1; 1].
(b) For the matrix [2 2; 2 -1] and λ = -2, we subtract λ from the diagonal entries to get [2-(-2) 2; 2 -1-(-2)] = [4 2; 2 1].
Solving the system of linear equations 4x + 2y = 0 and 2x + y = 0 gives us x = 0 and y = 0. Therefore, the eigenvector corresponding to λ = -2 is [0; 0].
(c) For the 3x3 matrix [3 1 -1; 0 2 0; 0 0 1] and λ = 3, we subtract λ from the diagonal entries to get [3-3 1 -1; 0 2-3 0; 0 0 1-3] = [0 1 -1; 0 -1 0; 0 0 -2].
Solving the system of linear equations given by this matrix, which is x + y - z = 0, -y = 0, and -2z = 0, gives us y = 0, z = 0, and x = 0. Therefore, the eigenvector corresponding to λ = 3 is [0; 0; 0].
Note: The eigenvectors in parts (b) and (c) are the zero vector, which is technically not considered an eigenvector. This suggests that the given eigenvalues for these matrices may be incorrect.
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