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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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1 -2
0 1
,0 3
3
1
-7
3
-1
7
7
10
7
10
(c) Add (-3) times the second row to the third row.
(d) Divide the third row by -4.
-2 3 | 7
1 -1 | -2
0 -4 | -4
-2 3 | 7
1 -1 | -2
0 1 1
The matrix obtained in this manner corresponds to the system of equations
*1 - 2v2
? 3v, =
7,
-2,
1,
(10)
which is equivalent to the original system (8). Note that the coefficients in Eqs. (10) form a triangular matrix. From the last of Eqs. (10) we have v3 = 1, from the second equation x2 = -2 + v3 = -1, and from the first equation x1 = 7 + 2v2 - 3v3 = 2. Thus we obtain
2^
x = |-1
which is the solution of the given system (8). Incidentally, since the solution is unique, we conclude that the coefficient matrix is nonsingular.
*2 - V3
Discuss solutions of the system
x1 - 2x2 + 3x3 = b1,
-x1 + x2 - 2x3 = b2, (11)
2x1 - x2 + 3x3 = b3
for various values of b1, 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
x-(•T)--(IMf (“
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 ith component of the vector x^), and let X = (xJ. Then Eq. (17) can be written as
(x
x\ 'cl +-------------------------+ x) 'c.
\xn] ci +••• + xr cJ Vm'ci +••• + xnn cn)
(n) \
c
n)
n
(xi1 ci + •" + xincn\
= Xc = 0.
(i8)
If det X = 0, then the only solution of Eq. (i8) is c = 0, but if det X = 0, there are nonzero solutions. Thus the set of vectors x (i),..., x( n) is linearly independent if and only if det X = 0.
I
EXAMPLE
3
Determine whether the vectors
i
<(i) = | 2
'2'
x(2) = | i i3,
x(3) =
—4'
ii
(i9)
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(x^) =
i
i
2 -4
3 -ii
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
cix(i) + c2x(2) + c3x(3) = 0.
Equation (20) can also be written in the form
1
2 i
2 -4 i i
3 -ii
0
= | 0 0
(20)
(2i)
and solved by means of elementary row operations starting from the augmented matrix
1 2-4 10^
i
i
-i
We proceed as in Examples i and 2.
3 -ii
(22)
(a) Add (-2) times the first row to the second row, and add the first row to the third row.
i 2 -4
0 -3 9
,0 5 —i5
i
362
Chapter 7. Systems ofFirst Order Linear Equations
(b) Divide the second row by -3; then add (—5) times the second row to the third row.
'1 2 —4 | 0'
0 1 —3 | 0
,0 0 0 10,
Thus we obtain the equivalent system
c1 + 2c2 - 4c3 = 0, c2 - 3c3 = 0.
(23)
From the second of Eqs. (23) we have c2 = 3c3, and from the first we obtain c1 =
4c3 - 2c2 = —2c3.
Thus we have solved for c1 and
c2 in terms of c3,
with the latter
remaining arbitrary. If we choose c3 = — 1 for convenience, then c1 = 2 and c2 = — 3. In this case the desired relation (20) becomes
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