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Department of Mathematics Linear Algebra and Probability
Homework Key
Assignment 8
April 18
- Find the determinant of each of the following matrices.
A =
[
]
B =
C =
(a) det(A) = 3 · 3 − (−1)(−1) = 10 (b) det(B) = (−1)(2)(−2) = 4 (c)
det(C) = det
R 3 − = 3R 1
R 2 − = 2R 1
det
R 2 ↔ R 3
− det
= R^3 −^ = 4R^2
− det
- Find the eigenvalues and corresponding eigenvectors of the following matrices.
A =
[
]
B =
(a) To find the eigenvalues, we solve det(A − λI) = 0. This gives
det(A − λI) = det
[
2 − λ 1 4 − 1 − λ
]
= (2 − λ)(− 1 − λ) − 4 = (λ − 3)(λ + 2) = 0.
So A has eigenvalues λ = 3 and λ = − 2.
In order to find the eigenvector corresponding to λ = 3, we substitute λ into the matrix equation (A − λI)x = 0 and solve for x :
[ 2 − λ 1 4 − 1 − λ
]
[
]
[
] R 2 + = 4R 1
R 1 ∗ = − 1
[
]
So we can choose the eigenvector corresponding to the eigenvalue λ = 3 to be the special solution x = (1, 1). Note that we could also choose any nonzero multiple of the special solution. We can use a similar approach to find the eigenvector corresponding to the eigen- value −2 : [ 2 − λ 1 4 − 1 − λ
]
[
]
[
] R 2 − = R 1
R 1 ∗ = 1/ 4
[
]
So we can choose the eigenvector corresponding to the eigenvalue λ = −2 to be the special solution x = (− 1 / 4 , 1), but we could also choose any nonzero multiple of the special solution.
(b) Solving det(B − λI) = 0, gives
det(B − λI) = det
1 − λ 2 0 0 2 − λ 1 0 0 3 − λ
(^) = (1 − λ)(2 − λ)(3 − λ) = 0.
So the eigenvalues are λ = 1, λ = 2, and λ = 3. In order to find the eigenvector corresponding to λ = 1, we solve (B − I)x = 0 :
R 1 ∗ = 1/ 2
R 3 ∗ = 1/ 2
R 2 − = R 1
R 2 − = R 3
R 2 ↔ R 3
So we can choose the eigenvector corresponding to λ = 1 to be the special solution x = (1, 0 , 0). In order to find the eigenvector corresponding to λ = 2, we solve (B − 2 I)x = 0 :
R 1 ∗ = − 1
R 3 − = R 2
