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Homework 6 Solutions - ALL ANSWERS ARE CORRECT - University of California, Los Angeles LIFESCIENC 30B $9.99   Add to cart

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Homework 6 Solutions - ALL ANSWERS ARE CORRECT - University of California, Los Angeles LIFESCIENC 30B

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Homework 6 Solutions Total points: 32 6.4.7: Show that M3 =  1 0 0 1 . Solution: M3 =  0 1 −1 −1   0 1 −1 −1   0 1 −1 −1  =  0 1 −1 −1  −1 −1 1 0  =  1 0 0 1 . 6.4.8: Using the point p =  5 0  as the test point, apply M three ti...

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Homework 6 Solutions

Total points: 32

February 2019


 
1 0
6.4.7: Show that M3 = .
0 1


Solution:
        
3 0 1 0 1 0 1 0 1 −1 −1 1 0
M = = = .
−1 −1 −1 −1 −1 −1 −1 −1 1 0 0 1

 
5
6.4.8: Using the point p = as the test point, apply M three times to calculate M p, M 2 p, and
0
M 3 p.


Solution:
 
0
Mp = ,
−5
 
2 −5
M p= ,
5
 
3 5
M p= .
0

   
2 −5 −4 2
6.4 FE 4: The matrix A = 0 3 2  has an eigenvector −2. What is its corresponding
0 −4 −3 4
eigenvalue?


Solution:       
2 −5 −4 2 −2 2
0 3 2   −2 =
  2  = (−1) −2 .

0 −4 −3 4 −4 4
Therefore,
λ = −1.

1

,6.4 FE 7: Compute the eigenvalues of the linear function
   
X 4X − 5Y
f( )=
Y 2X − 2Y



Solution: The matrix representation of this linear function is
 
4 −5
.
2 −2

The characteristic equation is
λ2 − 2λ + 2 = 0.
Therefore,
λ = 1 ± i.

 
−9 −8
6.4 FE 8: One of the eigenvalues of the matrix is 3. What is a corresponding eigen-
12 11
vector for it?

   
X −9 −8
Solution: Let’s denote by an arbitrary eigenvector with eigenvalue 3 of , so that
Y 12 11
    
−9 −8 X X
=3· .
12 11 Y Y

This gives you a system of two linear equations in X and Y , and you can solve the first or the
second equation to find that
3
Y = − X.
2
So, the vectors of the form  
X
,
− 32 X
where X 6= 0, are the eigenvectors with eigenvalue 3 of this matrix. Letting X = 2, you can choose
as your particular eigenvector  
2
V= .
−3




2

,  
4 5 −3
6.4 FE 9: One of the eigenvalues of the matrix 4 6 −4, is 2. What is a corresponding
8 11 −7
eigenvector for it?


Solution: Let’s denote by  
X
Y 
Z
an arbitrary eigenvector with eigenvalue 2 of this matrix, so that
    
4 5 −3 X X
4 6 −4   Y = 2 · Y .
 
8 11 −7 Z Z

This gives you a system of three linear equations in X, Y, and Z, namely
4X + 5Y − 3Z = 2X
4X + 6Y − 4Z = 2Y
8X + 11Y − 7Z = 2Z
After simplification, you will find that
X = 2Y, Z = 3Y
(to get started, look at the coefficients of X). So, the vectors of the form
 
2Y
 Y ,
3Y
where Y 6= 0, are the eigenvectors with eigenvalue 2 of this matrix. Letting Y = 1, you can choose
as your eigenvector  
2
V = 1 .

3

6.4 FE 10:

a) Solve for a and b in the equation
     
2 −3 9
a +b =
5 1 14
 
9
b) Use your answer to part (a) to give the coordinates of with respect to the basis
14
   
2 −3
, .
5 1



3

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