Answers to Selected Problems in Multivariable Calculus with Linear Algebra and Series

94 Pages·1995·1.974 MB·English

by  William F. Trench and Bernard Kolman (Auth.)

Answers to Selected Problems in
Multivariable Calculus
with Linear Algebra
and Series
WILLIAM F. TRENCH AND BERNARD KOLMAN
Drexel University
Section 1.1, page 21
2. (a) x = -1, y = 4 (b) x = -2, y = 2, z = 3
4. (a) x = 0, y = 3 (b) x = 3, y = 1
-2 0 2 3-2 4 6
6. (a) (b) (c)
-2 2 -5 10 4 3 3
(d) v indef ined (e) undefined
10. (a) 6 5 (b) -55
12. (a) c 1, c, 5, c = 10
1 = 5' C2 4
(b) r '■L = 3, r 2 3, r3 = 12, r4 = 3
(c) 21 (d) 21
13 6 5
14. (a) (b) undefined
13 2 1
49 18 5
(c) undefined (d) (e) undefined
39 30 3
— — r n
3 LI X 5 1 1 -2
4
20. (a) (b)
y
3 3j L _ 8 1 - 10
X
1 2 0 1 4
y
2 -1 1 0 = -2 1
z
0 1 -2 2 6
w *—- —J
1
T-1. (a.. +b..) + c. . = a.. + (b.. + c. .).
ij iJ ij iJ iJ iJ
T-2. [a..] + [-a..] = [a.. - a..] = [0].
ij ij iJ iJ
T-3. (a) r[sa ] = [r(sa )] = [(rs)a ] = rs[a ].
(b) (r + s)[a ] = [(r + s)a ] - [ra + sa^]
= [ra..] + [sa..] = r[a..] + s[a..].
ij ij iJ ij
(c) r([a + b ]) = r([a ] + [b ])
=r[a..] r[b..].
+
T-4. Write (a) S = 1 + 2 +··· + n and (b) S = n + (n-1)
n n
+···+ 1. Then add (a) and (b) to obtain
2S = (n + 1) + (n + 1) +···+ (n + 1) , whlee re (n+1)
n
n(n + 1)
appears n times on the right. Hence S
T-5. (a.. + b-)c_ + ···+ (a + b )c = a.c. + b-C,
i 11 n nn 11 11
+ · · · + a c + b c = (a-c. +· · ·+ a c )
nn nn 11 nn
+ (b-C- +···+ b c ); and
11 n n
(ca_ +···+ ca ) = c(a- + ··· + a ).
1 n 1 n
T-6. (a) ^aik(b + V - Z_/iAj + /_f±^\y
kj
k=l k=l k=l
P _p^ p
(b) ^ ( d . + e )c . =)^d c . +2_,e. c ;
k ik k ik k k kj
k=l k=l k=l
2
P
<°> Σ·<θ ■ <I>^) -Ζ(
taikb
kj
k=l k=l k=l
T T T
T-7. (a) (a!.)1 = at. = a..
ij Ji iJ
T
(b) Let c. = a.. + b..; then c.. = c. = a.. + b..
1 1 1
iJ iJ iJ iJ J J J
= aT.. +M bu. T . .
T T
(c) Let d.. = ra..; then d.. = d.. = ra.. = ra...
ij ■ ij ij ji ji ij
T T
T-8. If A = -A, then a.. = -a..; hence a.. = -a...
In particular, a.. = -a.., and therefore a.. = 0.
^ 11 11 ii
T-9. If AT = -A and AT = A, then A = -A; hence 2A = 0,
and A = 0.
T T
A + A A - A
T-10. B = — , C = —- . For uniqueness, let
(i) A = B + C where B^ = B and C^ = -C ;
±9 ±
Then (ii) AT = B^ + C^ = B - C . Adding (i) and
(ii) yields B = B; subtracting (ii) from (i) yields
ci - c·
T-ll. Define I = [e..] where e = 1 and e = 0
m ij ii ij
(1 < i < j < m) . Let I A = [b..]; then
b = ) e,a,.=e..a..=a... Similar proof for
ij /_j ik kj ii ij ij
k=l
3
AI = A.
n
T-12.. (a) Suppose the i-th row of A consists entirely of
zeros. Let C = AB; then
= bk = b = 0 (1 j ln)
^ 2_> J / °' v -
(b) Similar argument.
T T
T-13. (A ) = A, from Thm. 1.5(a). Since A is symmetric,
.
rp rp rp rn rp
A = A . Therefore (A ) = A ; hence A is
symmetric.
T-14.. Let a.. = b.. = 0 if i Φ j. If C = A + B, then
ij ij
c..=a..+b.. =0 + 0 = 0 if i^j. If D = AB,
then d.. = / a.-b. . = a..b.. = 0 if i φ j.
ij / lk kj il ij
T-15.. (a) (A + B)T = AT + BT = A -f B. (b) (AB)T = BTAT,
T T
from Thm. 1.5(d); since A = A and B = B,
(AB)T = BA.
T-16.. Let A and B be upper triangular. If C = A + B, then
c..=a..+b..=0+0=0 if i>i. If D = AB,
iJ ij ij J
then d.. = > a b- . = ) a.-b, .. If i > j , then
ij ^ #l1k kj £^ ik kj J
k>i
k > i for all terms in the last sum; hence b ,. = 0
for all these terms. Therefore, d.. = 0 if i > i.
4
Section 1.2, page 42
1 2 1 1 0 0
0 1 2 0 1 0
2. (a) (b)
0 0 1 0 0 1
0 0 0 0 0 Oj
4. A, C, D, and F are in row echelon form; D and F are in
reduced row echelon form.
6. (a) no solutions. (b) x 5 - J w, 5 w,
8
z = J - J w, w arbitrary (c) x = -1, y = 0, z
8. (i) a Φ ±3 (ii) a = -3 (iii) a = 3
10. (i) a + ±1 (ii) a = -1 (iii) a = 1
12. (a) Χη — 1} X« _ J- J X« _ i. J X . ~ L·
1 2 _ 11
(b) = ^x x = 0
Xl 3 3 V X2 6
x, arbitrary
14. (a) x^^ = x = x = 0. (b)
2 3 Xl 18 X4' X2 18 4'
x = — x , x arbitrary.
3 4 4
T-l. Multiplication of a row of [A 1 Y] by a nonzero
constant corresponds to multiplication of an
equation in the system AX = Y by the same constant,
This does not change the solutions of the system.
Similar argument applies to the other elementary
operations.
5
T-2. A row in the augmented matrix of a system in
n unknowns, with zeros in the first n columns and
a 1 in the (n + 1) -st corresponds to the equation
0·χ + + 0·χ = 1,
n
which has no solution.
For the converse, let [A|Y] be row equivalent
to [B|Z], which is in row echelon form. Since B
has no row with a "leading" 1 in the (n + l)-st
column, it follows (with the notation of Def. 2.3)
that j- < j < · ·· < j,· Hence, for i = 1, 2,...,k,
?
the i-th equation of BX = Z can be solved for x. in
terms of the remaining n - k unknowns, which can be
specified arbitrarily.
T-3. By definition of the elementary row operations, the
system BX = 0 is obtained by performing on AX = 0
operations which do not change the solutions of the
latter.
T-4. Suppose ad - be ^ 0, b^O, d^O. Then the
following matrices are row equivalent :
a b a d bd ad-bc 0
9
c d be bd be bd
-J *—
•1
1 1 0 1 0
J
be I 0 bd 0 1
Suppose ad - be ^ 0 and b = 0. Then d φ 0 and
a φ 0. The following matrices are row equivalent:
6
a 0 1 0 1 0
>
c d 0 d 0 1
A similar argument disposes of the case where d = 0.
For the converse, suppose ad - be = 0, but b
and d are nonzero. Then the following matrices are
row equivalent.
a b ad bd ad-bc 0 0 0
> 9
c d be bd_ - bc bd J>c bd
which is not row equivalent to I . If ad - bc = 0
and b = 0, then a = 0 or d = 0. Then
0 0 a b
J
c d c d
neither of which is row equivalent to I
r
similar argument disposes of the case where d = 0.
a b
T-5. If ad - bc φ 0, then A is row equivalent
c d
to I (Ex. T-4); hence Thm. 2.2 implies that AX = 0
and IX = 0 have the same solutions. Since the
?
latter has only the trivial solution, so does the
former.
If ad - bc = 0, then A is row equivalent to
1 0 0 0
B or C = , since these are the
0 0 0 0
7
only 2 x2 matrices, other than I«, in reduced row
echelon form. Since BX = 0 and CX = 0 have
nontrivial solutions, so does AX = 0, again by
Thm. 2.2.
T-6. Let A be an m x n matrix and define
0
E2 = • En =
0 1
^- .J
If a., φ 0 for some i and 1, then E. is not a
iJ J
solution of AX = 0; since 0·Ε. = 0, A and 0 are not
J
row equivalent, by Thm. 2.2.
Section 1.3, page 59
1 1 1
-2" 1 2-1 0-1
2 2
1 1 1 3 4 2 10
4. -2" 6. (a)
2 2
6 -9 -15 6 -3
1 1 1
-2"
2 2
1 2-1 0-1 1 2-1 0-1
(b) - 2 3 5 - 21 (c) 3 4 2 10
3 4 2 10 0 7 3-2-1
8
10 0 1 0 -4 10 0
8. 0 0 1 0 10 0 5 0
0 10 0 0 1 0 0 1
5_ 1
12 4
Ì Ì _ 1 1
10. (a) I ; (b) I ;
3 3 18 6
2 2 2 1 1
-1
2 2
0 - 13
i I
(c) singular
■i o-if'
-5 3
0 ! if
12. (a) ; singular (b) 1^
2 -1
o J
2 λ
1 -
5 5
(c) I ; -2
3
1 0
1 - 1 51 J
9