# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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Ê Ã J

132

Section 7.3

18.

21.

´ 1 -1ë / \ ' 0ë (2) ' 1 N

x1 Hence x =

= , or x1 - x2 = 0.

V 3 -3, x2 V 0 , I1

2

, or

a multiple thereof.

Since a12 = a21, the given matrix is Hermitian and we

know in advance that its eigenvalues are real. To find the eigenvalues and eigenvectors we must solve

'1-À, i 4 - i 1-À

x2

2

0

The determinant of coefficients

is (1-À)2 - i(-i) = X2 - 2X, so the eigenvalues are À1 = 0 and X2 = 2; observe that they are indeed real even though the given matrix has imaginary entries. The eigenvector corresponding to X1 must satisfy

1 i

-i 1

x

2

0

0

, or x1 + ix2 = 0. Note that the second

equation -ix1 + x2 = 0 is a multiple of the first. If x1 = 1, then x2 = i, and the eigenvector is

.(1)

1

(2)

1

. In a similar way we find that the eigenvector associated with X2 is x(

The eigenvalues and eigenvectors satisfy

' 1-À 0 0 ë / \ ' 0 ^

x1

2 1-À -2 x2 = 0

v 3 2 1-À , x3 V 0 ,

3

The determinant of coefficients is

(1-À)[(1-À) + 4] = 0, which has roots X = 1, 1 ± 2i. For X = 1, we then have 2x1 - 2x3 = 0 and 3x1 + 2x2 = 0. Choosing

x1 = 2 then yields

2

-3

as the eigenvector corresponding to

X = 1. For X = 1 + 2i we have

2x3 = 0 and 3x! + 2x2 - 2ix3 = 0,

-2ix! = 0, 2x! - 2ix2

0

yielding x1 = 0 and x3 = -ix2. Thus

is the eigenvector

corresponding to X = 1 + 2i. A similar calculation shows that

0

x

1

Section 7.3

133

0

is the eigenvector corresponding to X = 1 - 2i.

24. Since the given matrix is real and symmetric, we know that the eigenvalues are real. Further, even if there are repeated eigenvalues, there will be a full set of three linearly independent eigenvectors. To find the eigenvalues and eigenvectors we must solve

The determinant of

coefficients is (3-X)[-X(3-X)-4] - 2[2(3-X) -8] + 4[4+4X] = -X3 + 6X2 + 15X + 8. Setting this equal to zero and solving we find X1 = X2 = -1, X3 = 8. The eigenvectors

corresponding to X1 and X2 must satisfy

' 3-X 2 4 " / \ ' 0 N

x1

2 -X 2 x2 = 0

V 4 2 3-Xy x3 V 0 ,

3

2 4" / \ ' 0 ë

4 x1

2 1 2 x2 = 0 ; hence there is only the single

4 4 ó x3 V 0 ,

2 3

relation 2x1 + 2x3 = 0 to be satisfied.

2

x

+

Consequently, two of the variables can be selected arbitrarily and the third is then determined by this equation. For example, if x1 = 1 and x3 = 1, then x2 =

1

(1)

4, and we obtain the eigenvector x if x1 = 1 and x2 = 0, then x3 = -1, and we have the

(2)

Similarly,

1

eigenvector x

(1)

0

, which is linearly independent of

x . There are many other choices that could have been made; however, by Eq.(38) there can be no more than two linearly independent eigenvectors corresponding to the eigenvalue -1. To find the eigenvector corresponding to X3 we must solve

134

Section 7.4

m 2 4" / \ ' 0 N

- x1

2 -8 2 x2 = 0

V 4 2 -5, x3 V 0 ,

3

Interchange the first and

second rows and use row reduction to obtain the equivalent system x1 - 4x2 + x3 = 0, 2x2 - x3 = 0. Since there are two equations to satisfy only one variable can be assigned an arbitrary value. If we let x2 = 1, then

2

x3 = 2 and x1 = 2, so we find that x

(3)

27. We are given that Ax = b has solutions and thus we have (Ax,y) = (b,y). From Problem 26, though,

(Ax,y)= (x, A*y) = 0. Thus (b,y) = 0. For Example 2,

A* = A =

´ 1 -1 2 4

-2 1 -1 V 3 -2 3Ó

and, using row reduction, the augmented

and

' 1 -1 2 0ë ' 1ë

matrix for A*y = 0 becomes 0 1- 3 0 . Thus y = c 3

V 0 0 0 0, 1I

hence (b,y) = b1 + 3b2 + b3 = 0.

1

Section 7.4, Page 371

Use Mathematical Induction. It has already been proven that if x(1) and x(2) are solutions, then so is

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