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Title: TMC1874 Mathematics For Computing tutorial 7 solution
Description: Tutorial Solution Chapter 7: Eigenvalues and Eigenvectors
Description: Tutorial Solution Chapter 7: Eigenvalues and Eigenvectors
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TMC1874 Mathematics For Computing
Tutorial Solution
Chapter 7: Eigenvalues and Eigenvectors
1
...
Then the characteristics equation is
|A − λI| =
−11 − λ
18
−6
10 − λ
=
0,
(−11 − λ)(10 − λ) + 108 = 0,
λ2 + λ − 2
=
0
...
To find the eigenvectors corresponding to λ = −2, we solve (A − λ I)x = 0
...
K
...
−1
The eigenvalues are 1, 2, 4
...
1
1
The eigenvector corresponding to λ = 2 is x2 = 0
...
2
(c) The characteristics equation is
λ3 − 5λ2 + 2λ + 8 = 0
...
−8
The eigenvector corresponding to λ = −1 is x1 = 10
...
1
1
The eigenvector corresponding to λ = 4 is x3 = 0
...
The eigenvalues are 1, 3, 4
...
K
...
2
1
0
...
1
Page 2 of 3
TMC1874 Mathematics For Computing
2
...
Thus there is one eigenvalue
of multiplicity 3: λ = 4
...
Since we can only find one linearly
0
independent eigenvector
...
(b) First, find the eigenvalues, i
...
λ = −4, 3
...
(c) The eigenvalues are λ = 1, 2, 3
...
(d) The eigenvalues are λ = −1, 2, 2, 5
...
Since
3
0
we can only find one linearly independent eigenvector corresponding to λ2 = λ3 = 2
...
3
...
0
−1
1
Solution
−2
Let D = 0
0
0
−2
0
0
1
0 and P = 1
3
0
1
0
−1
1
1
...
W
...
Tiong
Page 3 of 3
Title: TMC1874 Mathematics For Computing tutorial 7 solution
Description: Tutorial Solution Chapter 7: Eigenvalues and Eigenvectors
Description: Tutorial Solution Chapter 7: Eigenvalues and Eigenvectors