Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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3. Form the augmented matrix and use row reduction.
' 1 2 1 2 N

2 1 1 . 1
V 1 1 2 . 1
Add (2) times the first row to the second and add (1) times the first row to the third.
' 1 2 1 2 N

0 3 3 3
 
V 0 3 3 3 ,

Add (1) times the second row to the third row and then multiply the second row by (1/3).
' 1 2 1 2 N

0 1 1 . 1
V 0 0 0 . 0,
Sectuib 7.3
129
Since the last row has only zero entries, it may be
dropped. The second row corresponds to the equation
x2  x3 = 1. We can assign an arbitrary value to either x2
or x3 and use this equation to solve for the other. For
example, let x3 = c, where c is arbitrary. Then
x2 = 1 + c. The first row corresponds to the equation
xi + 2x2  x3 = 2, so
x1 = 2  2x2 + x3 = 2  2(1+c)+c = c.
6. To determine whether the given set of vectors is linearly independent we must solve the system
c1x(1) + c2x(2) + c3x(3) = 0 for c1, c2, and c3. Writing this in scalar form, we have c1 + c3 = 0
c1 + c2 = 0, so the
c2 + c3 = 0
augmented matrix is
1 1 0
0 1 1
Row reduction yields
1 0 1
1 1
From the third row we have c3 = 0. Then from the second row, c2  c3 = 0, so c2 = 0. Finally from the first row
c1 + c3 = 0, so c1 = 0. Since c1 = c2 = c3 = 0, we
conclude that the given vectors are linearly independent.
1
0
1
0
0
0
0
0
0
0
0
2
0
8. As in Problem 6 we wish to solve the system
(1) (2) (3) (4) ë p ^
c1x + c2x + c3x + c4x = 0 for c1, c2, c3, and
c4. Form the augmented matrix and use row reduction.
130
Section 7.3
14.
' 1 1 2 3 0 ë
  
2 0 1 0 . 0
2 3 1 1 . 0
V 3 1 0 3 . 0,
Add (2) times the first row to the second, add (2) times the first row to the third, and add (3) times the first row to the fourth.
' 1 1 2 3 . 0 ^
0 2 3 6 . 0
0 5 5 5 . 0
V 0 4 6 12 . 0î
Multiply the second y
X3
w
o
r
the second row to the third
row to the fourth.
' 1 1 2 3 . 0 ^
0 1 3/2 3 . 0
0 0 5/2 0
0
1

V 0 0 0 0
0
The third row is equivalent to the equation c3 + 4c4 = 0. One way to satisfy this equation is by choosing c4 = 1; then c3 = 4. From the second row we then have c2 =  (3/2)c3  3c4 =  6 + 3 = 3. Then, from the first row, c1 = c2 + 2c3 + 3c4 = 3 + 8  3 = 2. Hence the given vectors are linearly dependent, and satisfy
2x(1)  3x(2) + 4x(3)  x(4) = 0.
Let t = t0 be a fixed value of t in the interval
0 < t < 1. To determine whether x(1) (t0) and x(2) (t0) are
Section 7.3
ãçã
Ã5.
linearly dependent we must solve crx<r)(tr)+c2x<2)(tr}=0. We have the augmented matrix
tre
tr
ã
Multiply the first row by (tr) and add to the second row
to obtain
ã
ã
ã
Ã ó
Thus, for example, we can choose cr = Ã and c2 = e ã, and hence the given vectors are linearly dependent at tr. Since tr is arbitrary the vectors are linearly dependent at each point in the interval. However, there is no linear relation between x(r) and x(2) that is valid throughout the interval Ã < t < Ã. For example, if tr Ô tr, and if cr and c2 are chosen as above, then
crx(r)(tr) + c2x(2)(tr)
e + ’ã ' ã ë N ô Ã1
ã e Ã e
ó ã
1
e
Ã
ó
tretr  ’ã , ’ãº’ã  ’ãº’ã  Ã ó
ã ãã
Hence the given vectors must be linearly independent on Ã < t < Ã. In fact, the same argument applies to any interval.
To find the eigenvalues and eigenvectors of the given
matrix we must solve
5Ë, Ã 3 ÃË
x.
2
The
determinant of coefficients is (5Ë) (ÃË)  (Ã)(3) = Ã,
or Ë2  6Ë + 8 = Ã. Hence Ëã = 2 and Ë2 = 4 are the eigenvalues. The eigenvector corresponding to Ëã must
satisfy
ëã
x2
2
"Ã4
, or 3xr  x2 = Ã. If we let
xr = Ã, then x2 = 3 and the eigenvector is x
(ã)
or
any constant multiple of this vector. Similarly, the eigenvector corresponding to Ë2 must satisfy
t
Ã
Ã
e
t
Ã
e
x
ã