Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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2 A + B = 1 4 + 5 2 + 0 = 5 9 2

6
v4 + 6 2 + 1 6+2, v 2 3 8 ,
1c. Using Eq.(9) and following Example 1 we have
AB
4 + 2 + 0
12  2  6
8  1 + 18
2  10 + 0 6 + 10  1 4 + 5 + 3
(AB)C = A(BC) =
7 11
11 20 17
3 12
3 + 0 + 0
9 + 0  2
6 + 0 + 6
which yields the correct answer.
' 6 5 7" ' 5 3 3 N
6. AB = 1 9 1 and BC = 1 7 3
v1 2 8, V 2 3 2 ,
so that
In problems 10 through 19 the method of row reduction, as illustrated in Example 2, can be used to find the inverse matrix or else to show that none exists. We start with the original matrix augmented by the indentity matrix, describe a suitable sequence of elementary row operations, and show the result of applying these operations.
10. Start with the given matrix augmented by the identity
matrix.
4 . 1
0
Add 2 times the first row to the second row.
1
4
1
0
0 11 . 2 1 V Ó
Multiply the second row by (1/11).
4 2 6 î
1
2 3 . 0 1 î
Section 7.2
125
' 14. 1 0 ^
v0 1 . 2/11 1/11,
Add (4) times the second row to the first row.
'1 0 . 3/11 4/11^
v0 1 . 2/11 1/11 ,
Since we have performed the same operation on the given matrix and the identity matrix, the 2 x 2 matric appearing on the right side of this augmented matrix is the desired inverse matrix. The answer can be checked by multiplying it by the given matrix; the result should be the indentity matrix.
12. The augmented matrix in this case is:
' 1 2 3 1 0 0 ë
2 4 5 0 1 0
V 3 5 6 0 0 1,
Add (2) times the first row to the second row and (3) times the first row to the third row.
' 1 2 3 1 0 0 ë
0 0 1 2 1 0
V 0 1 3 3 0 1 ,
Multiply the second and third rows by (1) and interchange them.
' 1 2 3 1 0 0 ë
0 1 3 3 0 1
0 0 1 2 1 0
V /
Add (3) times the third row to the first and second
126
Section 7.2
' 1 2 0 5 3 0
rows. 0 1 0 .3 3 1
v 0 0 1 . 2 1 0
Add (2) times the second row to th
' 1 0 0 1 3 2 N
0 1 0 3 3 1
0 0 1 2 1 0
V )
The desired answer appears on the r
augmented matrix.
14. Again, start with the given matrix augmented by the
"1 2 1 . 1 0 0 "
identity matrix.
2 1 8 . 0 1 0
Add (2) times the first row to the second row and add(1) times the first row to the third row.
' 1 1 1 0 0ë
2
0 5 10 . 2 1 0
V 0 1 1 0 1,
1
CO
the second row to the third
' 1 1 1 0 0
2
0 5 10 . 2 1 0
v0 0 0 .3/5 4/5 0 ó
Since the third row of the left matrix is all zeros, no further reduction can be performed, and the given matrix is singular.
^1 2 7 . 0 0 1)
Section 7.3
127
22. x' =
2e
2t
2t
; and
' 3 2 ' CN ' 4 ' 2t ' 124" 2t ' 8 N

m
x = e = e =
v2 2, V2 2, V 2 Ó 8 V 4 ,
³
2t
´ n 3t _ 2t4
3e 2e
V12e 3t 2e2ty
—3t
ee
—3t
2t
—4e
2t
e
Section 7.3, Page 366
1. Form the augmented matrix, as in Example 1, and use row reduction.
' 1 0 1 0 ë

VV 3 1 1 1
1 1 2 2 ,

Add (3) times the first row to the second and add the first row to the third.
' 1 0 1 . 0 ë
VV 0 1 4 . 1
V 0 1 1 . 2 ,
Add (1) times the second row to the third.
' 1 0 1 . 0 ë
VV 0 1 4 . 1
V 0 0 3 . 1,
The third row is equivalent to  3x3 = 1 or x3 =  1/3. Likewise the second row is equivalent to x2 + 4x3 = 1, so x2 = 7/3. Finally, from the first row, x1  x3 = 0, so x1 =  1/3. The answer can be checked by substituting into the original equations.
8
4
È —2 )
128
Section 7.3
2 . The augmented matrix is
' 1 2 1 1 N
2 1 1 1
V 1 1 2 1
1
Row reduction then
' 1 2 1 . 1ë
yields 0 3 3 . 1
v 0 0 0 . 1,
The last row corresponds to the equation
0x1 + 0x2 + 0x3 = 1, and there is no choice of x1, x2, and x3 that satisfies this equation. Hence the given system of equations has no solution.