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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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for various values of bj, b2, and b3.
360
Chapter 7. Systems of First Order Linear Equations
Observe that the coefficients in the system (11) are the same as those in the system (8) except for the coefficient of x3 in the third equation. The augmented matrix for the system (11) is
1 -2 3 | b1
-1 1 -2 | b2| . (12)
2 -1 3 | b3)
By performing steps (a), (b), and (c) as in Example 1 we transform the matrix (12) into
'1 -2 3 | b1 \
0 1 -1 | -b1 - b2 I . (13)
y0 0 0 | b1 + 3 b2 + b3J
The equation corresponding to the third row of the matrix (13) is
b1 + 3b2 + b3 = 0; (14)
thus the system (11) has no solution unless the condition (14) is satisfied by b1, b2, and b3. It is possible to show that this condition is just Eq. (5) for the system (11).
Let us now assume that b1 = 2, b2 = 1, and b3 = 5, in which case Eq. (14) is
satisfied. Then the first two rows of the matrix (13) correspond to the equations
x1 - 2x2 + 3x3 = 2,
x2 - x3 = -3.
(15)
To solve the system (15) we can choose one of the unknowns arbitrarily and then solve for the other two. Letting x3 = a, where a is arbitrary, it then follows that
x2 = a 3,
x1 = 2 (a 3) 3a + 2 = a 4.
If we write the solution in vector notation, we have
-C?) 4 i)+(:i)- (1*
It is easy to verify that the second term on the right side of Eq. (16) is a solution of the nonhomogeneous system (11), while the first term is the most general solution of the homogeneous system corresponding to (11).
Row reduction is also useful in solving homogeneous systems, and systems in which the number of equations is different from the number of unknowns.
Linear Independence. A set of k vectors x(1),..., x(k) is said to be linearly dependent if there exists a set of (complex) numbers c1t..., ck, at least one of which is nonzero, such that
c1x(1) + + ckx(k) 0. (17)
In other words x(1),..., x(k) are linearly dependent if there is a linear relation among them. On the other hand, if the only set c1t..., ck for which Eq. (17) is satisfied is c1 c2 ck 0, then x(1),..., x(k) are said to be linearly independent.
7.3 Systems of Linear Algebraic Equations; Linear Independence, Eigenvalues, Eigenvectors 361
Consider now a set of n vectors, each of which has n components. Let xiJ = x\j) be the i th component of the vector x(j), and let X = (Xi) Then Eq. (17) can be written as
(x (1
Vxn1) c1 + + xnn> cj Vxn1 c1 + + xnn cn)
(n)
c
(n)
n
(X11 c1 + " + X1ncn\
= Xc = 0.
(18)
If det X = 0, then the only solution of Eq. (18) is c = 0, but if det X = 0, there are nonzero solutions. Thus the set of vectors x(1),..., x(n) is linearly independent if and only if det X = 0.
I
EXAMPLE
3
Determine whether the vectors
1
t(1) = I 2
2
x(2) = I 1
3
x(3)
4
11
(19)
are linearly independent or linearly dependent. If linearly dependent, find a linear relation among them.
To determine whether x(1), x(2), and x(3) are linearly dependent we compute det(xij), whose columns are the components of x(1), x(2), and x(3), respectively. Thus
det(xiJ) =
1
1
2 -4
3 -11
and an elementary calculation shows that it is zero. Thus x(1), x(2), and x(3) are linearly dependent, and there are constants c1, c2, and c3 such that
c1x(1) + c2x(2) + c3x(3) = 0.
Equation (20) can also be written in the form
(20)
1
2
1
2 -4 1 1
3 -11
0
= I 0 0
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