A= 4 1 , -30 -7 with eigenvalue = -2
the answer is = 1 , -6
I am following the directions but I am not getting this!! Can someone please help!
the answer is = 1 , -6
I am following the directions but I am not getting this!! Can someone please help!
-
λI - A =
(λ - 4) (-1)
(30) (λ + 7)
When the eigenvalue λ = -2, we have λI - A =
(-2 - 4) (-1)
(30) (-2 + 7)
=
-6 -1
30 5
Adding 5 times the first row to the second row, we have:
-6 -1
0 0
Multiplying the first row by -1, we have:
6 1
0 0
Thus, we have:
6x + 1y = 0
Let x = s, where s is an arbitrary scalar.
y = -6x = -6s
We have:
[x y] = [s -6s] = s[1 -6]
Hence, [1 -6] is an eigenvector.
Note that eigenvectors are not unique.
In fact, there are infinitely many eigenvectors corresponding to the eigenvalue -2.
Thus, your answer is likely to be correct, though it is not [1 -6]
I believe you got [(-1/6) 1] as your eigenvector. That is fine as well.
(λ - 4) (-1)
(30) (λ + 7)
When the eigenvalue λ = -2, we have λI - A =
(-2 - 4) (-1)
(30) (-2 + 7)
=
-6 -1
30 5
Adding 5 times the first row to the second row, we have:
-6 -1
0 0
Multiplying the first row by -1, we have:
6 1
0 0
Thus, we have:
6x + 1y = 0
Let x = s, where s is an arbitrary scalar.
y = -6x = -6s
We have:
[x y] = [s -6s] = s[1 -6]
Hence, [1 -6] is an eigenvector.
Note that eigenvectors are not unique.
In fact, there are infinitely many eigenvectors corresponding to the eigenvalue -2.
Thus, your answer is likely to be correct, though it is not [1 -6]
I believe you got [(-1/6) 1] as your eigenvector. That is fine as well.