MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS #1. Page 169; Exercise 2. Let A = 2 1 3 -1 2 0 3 -2 1 . (a) Find adj A. (b) Compute det(A). (c) Verify Theorem 3.12; that is, show that A (adj A) = (adj A) A = det(A) I 3 . Solution: (a) First we compute the cofactors of A. We have A 11 =(-1) 1+1 2 0 -2 1 =2 A 21 =(-1) 2+1 1 3 -2 1 = -7 A 31 =(-1) 3+1 1 3 2 0 = -6 A 12 =(-1) 1+2 -1 0 3 1 =1 A 22 =(-1) 2+2 2 3 3 1 = -7 A 32 =(-1) 3+2 2 3 -1 0 = -3 A 13 =(-1) 1+3 -1 2 3 -2 = -4 A 23 =(-1) 2+3 2 1 3 -2 =7 A 33 =(-1) 3+3 2 1 -1 2 =5 Then the adjoint of A is the matrix adj A = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 = 2 -7 -6 1 -7 -3 -4 7 5 (b) We can compute the determinant by expanding along the first row of A: det(A)= a 11 A 11 + a 12 A 12 + a 13 A 13 = (2) (2) + (1) (1) + (3) (-4) = -7 (c) We have the matrix products A (adj A)= 2 1 3 -1 2 0 3 -2 1 2 -7 -6 1 -7 -3 -4 7 5 = -7 0 0 0 -7 0 0 0 -7 = det(A) I 3 . (adj A) A = 2 -7 -6 1 -7 -3 -4 7 5 2 1 3 -1 2 0 3 -2 1 = -7 0 0 0 -7 0 0 0 -7 = det(A) I 3 . #2. Page 169; Exercise 4. Find the inverse of the matrix in Exercise 2 by the method given in Corollary 3.4. 1
Transcript
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
#1. Page 169; Exercise 2. Let A =
!
"2 1 3!1 2 0
3 !2 1
#
$.
(a) Find adjA.(b) Compute det(A).(c) Verify Theorem 3.12; that is, show that A (adjA) = (adjA) A = det(A) I3.
Solution: (a) First we compute the cofactors of A. We have
A11 = (!1)1+1
%%%%2 0!2 1
%%%% = 2
A21 = (!1)2+1
%%%%1 3!2 1
%%%% = !7
A31 = (!1)3+1
%%%%1 32 0
%%%% = !6
A12 = (!1)1+2
%%%%!1 0
3 1
%%%% = 1
A22 = (!1)2+2
%%%%2 33 1
%%%% = !7
A32 = (!1)3+2
%%%%2 3!1 0
%%%% = !3
A13 = (!1)1+3
%%%%!1 2
3 !2
%%%% = !4
A23 = (!1)2+3
%%%%2 13 !2
%%%% = 7
A33 = (!1)3+3
%%%%2 1!1 2
%%%% = 5
Then the adjoint of A is the matrix
adjA =
!
"A11 A21 A31
A12 A22 A32
A13 A23 A33
#
$ =
!
"2 !7 !61 !7 !3!4 7 5
#
$
(b) We can compute the determinant by expanding along the first row of A:
#2. Page 169; Exercise 4. Find the inverse of the matrix in Exercise 2 by the method given in Corollary3.4.
1
2 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Solution: The inverse of the matrix is
A!1 =1
det(A)(adjA) =
!
&"
! 27 1 6
7
! 17 1 3
7
47 !1 ! 5
7
#
'$
#3. Page 172; Exercise 1. If possible, solve the following linear systems by Cramer’s rule:
2 x1 + 4 x2 + 6x3 = 2x1 + 2x3 = 0
2 x1 + 3x2 ! x3 = !5
Solution: Denote the coe!cient matrix of the system by
A =
!
"2 4 61 0 22 3 !1
#
$ =" |A| = (2)%%%%
0 23 !1
%%%%! (4)%%%%
1 22 !1
%%%% + (6)%%%%
1 02 3
%%%% = 26.
Then we have the values
x1 =
%%%%%%
2 4 60 0 2!5 3 !1
%%%%%%|A| =
(2)%%%%
0 23 !1
%%%%! (4)%%%%
0 2!5 !1
%%%% + (6)%%%%
0 0!5 3
%%%%26
=!5226
= !2
x2 =
%%%%%%
2 2 61 0 22 !5 !1
%%%%%%|A| =
(2)%%%%
0 2!5 !1
%%%%! (2)%%%%
1 22 !1
%%%% + (6)%%%%
1 02 !5
%%%%26
=026
= 0
x3 =
%%%%%%
2 4 21 0 02 3 !5
%%%%%%|A| =
(2)%%%%
0 03 !5
%%%%! (4)%%%%
1 02 !5
%%%% + (2)%%%%
1 02 3
%%%%26
=2626
= 1
Hence the solution is x1 = !2, x2 = 0, and x3 = 1.
#4. Page 172; Exercise 3. Solve the following linear system for x3, by Cramer’s rule:
2 x1 + x2 + x3 = 63 x1 + 2x2 ! 2 x3 = !2
x1 + x2 + 2x3 = !4
Solution: Denote the coe!cient matrix of the system by
A =
!
"2 1 13 2 !21 1 2
#
$ =" |A| = (2)%%%%
2 !21 2
%%%%! (1)%%%%
3 !21 2
%%%% + (1)%%%%
3 21 1
%%%% = 5.
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS 3
Then we have the value
x3 =
%%%%%%
2 1 63 2 !21 1 !4
%%%%%%|A| =
(2)%%%%
2 !21 !4
%%%%! (1)%%%%
3 !21 !4
%%%% + (6)%%%%
3 21 1
%%%%5
=45
#5. Page 187; Exercise 2. Determine the head of the vector(!2
5
)whose tail is (!3, 2). Make a sketch.
Solution: Say that the tail is P (!3, 2) and the head is Q(x, y). Then we have the directed line segment
!!#PQ =
(x! (!3)
y ! 2
)=
(!2
5
).
Since x + 3 = !2 and y ! 2 = 5, we see that x = !5 and y = 7. Hence the head is (!5, 7). A sketch canbe found below.
y
• (!5,7) !7
!
!5
!
!3
•
!!!!!!!!!!!!!!!!!!!!!!!!!!
(!3,2)
!
!1
""|!5
| |!3
| |!1
|1
| |3
| |5
x
##
#6. Page 187; Exercise 5. For what values of a and b are the vectors(
a! b2
)and
(4
a + b
)equal?
Solution: Upon equating both components, we find the system of equations
a ! b = 4a + b = 2
Adding these two equations together gives 2 a = 6, so that a = 3. Similarly, subtracting the two equationsgives !2 b = 2, so that b = !1. Hence the values are a = 3 and b = !1.
4 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
#7. Page 187; Exercise 7. In Exercises 7 and 8, determine the components of each vector!!#PQ.
(a) P (1, 2), Q(3, 5)(b) P (!2, 2, 3), Q(!3, 5, 2)
Solution: (a) We form the directed line segment by subtracting:
P (1, 2)Q(3, 5)
*=" !!#
PQ =(
3! 15! 2
)=
(23
).
Hence the components are 2 and 3.(b) Again, we form the directed line segment by subtracting:
P (!2, 2, 3)Q(!3, 5, 2)
*=" !!#
PQ =
!
"(!3)! (!2)
5! 22! 3
#
$ =
!
"!1
3!1
#
$ .
Hence the components are !1, 3, and !1.
#8. Page 187; Exercise 12. Compute u + v, 2u! v, 3u! 2v, and 0! 3v if
(a) u =
!
"123
#
$, v =
!
"201
#
$;
(b) u =
!
"2!1
4
#
$, v =
!
"12!3
#
$;
(c) u =
!
"10!1
#
$, v =
!
"!1
14
#
$.
Solution: (a) We perform operations component-wise on u =+
1 2 3,T and v =
+2 0 1
,T :
u + v =
!
"324
#
$ , 2u! v =
!
"045
#
$ , 3u! 2v =
!
"!1
67
#
$ , 0! 3v =
!
"!6
0!3
#
$
(b) For u =+
2 !1 4,T and v =
+1 2 !3
,T :
u + v =
!
"311
#
$ , 2u! v =
!
"3!411
#
$ , 3u! 2v =
!
"4!718
#
$ , 0! 3v =
!
"!3!6
9
#
$
(c) For u =+
1 0 !1,T and v =
+!1 1 4
,T :
u + v =
!
"013
#
$ , 2u! v =
!
"3!1!6
#
$ , 3u! 2v =
!
"5!2!11
#
$ , 0! 3v =
!
"3!3!12
#
$
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS 5
#9. Page 187; Exercise 14. Let
x =(
12
), y =
(!3
4
), z =
(r4
), and u =
(!2
s
).
Find r and s so that(a) z = 2x,(b) 3
2 u = y,(c) z + u = x.
Solution: (a) We have the equation(
r4
)= z = 2x =
(24
)=" r = 2
(b) We have the equation(!332 s
)= 3
2 u = y =(!3
4
)=" 3
2 s = 4 =" s =83
(c) We have the equation(
r ! 24 + s
)= z + u = x =
(12
)=" r ! 2 = 1
s + 4 = 2 =" r = 3s = !2
#10. Page 187; Exercise 16. If possible, find scalars c1 and c2 so that
c1
(1!2
)+ c2
(3!4
)=
(!5
6
).
Solution: The left-hand side of the equation simplifies to give the equation(
c1 + 3 c2
!2 c1 ! 4 c2
)=
(!5
6
)=" c1 + 3 c2 = !5
!2 c1 ! 4 c2 = 6
Upon adding twice the first equation to the second equation, we see that 2 c2 = !4, so that c2 = !2.Substituting this back into the first equation, we see that c1 ! 6 = !5, so that c1 = 1. Hence the desiredscalars are c1 = 1 and c2 = !2.
#11. Page 187; Exercise 17. If possible, find scalars c1, c2, and c3 so that
c1
!
"12!3
#
$ + c2
!
"!1
11
#
$ + c3
!
"!1
41
#
$ =
!
"2!2
3
#
$ .
6 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Solution: The left-hand side of the equation simplifies to give the equation!
"c1 ! c2 ! c3
2 c1 + c2 + 4 c3
!3 c1 + c2 ! c3
#
$ =
!
"2!2
3
#
$ ="c1 ! c2 ! c3 = 2
2 c1 + c2 + 4 c3 = !2!3 c1 + c2 ! c3 = 3
We list the augmented matrix for this system, and its reduced row echelon form:!
"1 !1 !1 22 1 4 !2!3 1 !1 3
#
$ ="
!
"1 0 1 00 1 2 00 0 0 1
#
$
Hence the system is inconsistent, so it is impossible; the scalars c1, c2, and c3 do not exist.
#12. Page 196; Exercise 2. Let V be the set of all 2$2 matrices A =(
a bc d
)such that the product a b c d =
0. Let the operation % be standard addition of matrices and the operation & be standard multiplication ofmatrices.
(a) Is V closed under addition?(b) Is V closed under scalar multiplication?(c) What is the zero vector in the set V ?(d) Does every matrix A in V have a negative that is in V ? Explain.(e) Is V a vector space? Explain.
Solution: (a) No, it is not closed under addition. Consider the matrices
A =(
1 01 0
)and B =
(0 10 1
)=" A%B =
(1 11 1
).
Then A, B ' V yet A%B (' V .(b) Yes, it is closed under scalar multiplication. For any scalar r ' R, we have the matrix
A =(
a bc d
)=" r A =
(r a r br c r d
)=" (r a) (r b) (r c) (r d) = r4 (a b c d) = r4 · 0 = 0.
(c) The zero vector is O2 =(
0 00 0
).
(d) Yes, every matrix has a negative that is in V . For any A ' V , we can choose the negative!A = (!1) A.Since V is closed under scalar multiplication, this element is in V .(e) No, V is not a vector space. This is because V is not closed under %.
#13. Page 196; Exercise 8. In Exercises 7 through 11, the given set together with the given operations isnot a vector space. List the properties of Definition 4.4 that fail to hold: The set of all ordered pairs of realnumbers with the operations
(x, y)% (x", y") = (x + x", y + y")
andr & (x, y) = (x, r y) .
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS 7
Solution: All properties hold except for Property (6). If we denote If we denote scalars c, d ' R and vectorsu = (x, y), v = (x", , y"), and w = (x"", y""), then we have
#14. Page 197; Exercise 9. In Exercises 7 through 11, the given set together with the given operations isnot a vector space. List the properties of Definition 4.4 that fail to hold: The set of all ordered triples ofreal numbers with the operations
(x, y, z)% (x", y", z") = (x + x", y + y", z + z")
andr & (x, y, r) = (x, 1, z) .
Solution: All properties hold except for Properties (5), (6), and (8). If we denote scalars c, d ' R andvectors u = (x, y, z), v = (x", y", z"), and w = (x"", y"", z""), then we have
(1) u% v = (x + x", y + y", z + z") = (x" + x, y" + y, z" + z) = v% u.(2) u%(v%w) =
For (5), (6), and (8), we have the following counterexamples:(5) Let c = 1 and u = v = (1, 0, 0). Then c& (u% v) = (2, 1, 0), yet c& u% c& v = (2, 2, 0).(6) Let c = d = 1 and u = (1, 0, 0). Then (c + d)& (u% v) = (1, 1, 0), yet c& u% d& v = (2, 2, 0).(8) Let u = (1, 0, 0). Then 1& u = (1, 1, 0) is di"erent from u.
#15. Page 197; Exercise 10. In Exercises 7 through 11, the given set together with the given operations is
not a vector space. List the properties of Definition 4.4 that fail to hold: The set of all 2$1 matrices(
xy
),
where x ) 0, with the usual operations in R2.
Solution: All properties hold except for Properties (4) and (b). The set V is closed under addition because
if we choose the vectors u =(
xy
)and v =
(x"
y"
)for some x ) 0 and x" ) 0, then x + x" ) 0 so that
u % v ' V as well. Properties (1) and (2) hold because % is the usual operation for R2. Property (3)
8 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
holds because the 2 $ 1 zero matrix has a nonpositive x-coordinate. For (4) and (b), denote the scalar
c = !1 and the vector u =(!1
0
). It is a vector in V because it has negative x-coordinate. However
c& u = !u =(
10
)is not in V because it has positive x-coordinate. Hence V does not contain negatives,
and it is not closed under scalar multiplication.
#16. Page 197; Exercise 12. Let V be the set of all positive real numbers; define % by u % v = uv (% isordinary multiplication) and define & by c& v = vc. Prove that V is a vector space.
Solution: V is closed under % and & because both uv and vc are positive real numbers. In order tocheck that V is a vector space, we verify eight properties. For scalars c, d ' R and positive real numbersu, v, w ' V , we have
(1) u% v = uv = vu = v% u.(2) u% (v%w) = u (vw) = (uv)w = (u% v)%w.(3) Denote 0 = 1. This is a positive real number, so 0 ' V . Moreover, u% 0 = 0% u = u 1 = u.(4) Denote !u = (1/u). This is a positive real number, so !u ' V . Moreover, u % !u = !u % u =
u (1/u) = 1 = 0.(5) c& (u% v) = c& (uv) = (uv)c = uc vc = c& u% c& v.(6) (c + d)& u = uc+d = uc ud = c& u% d& u.(7) c& (d& u) = c& ud = uc d = (c d)& u.(8) 1& u = u1 = u.
Hence, according to Definition 4.4, we see that V is a vector space.
#17. Page 205; Exercise 2. Let W be the set of all points in R3 that lie in the x y-plane. Is W a subspaceof R3? Explain.
Solution: Yes, W is a subspace of R3. Let us write W in the form
W =
01
2x =
!
"xyz
#
$ ' R3
%%%% z = 0
34
5 .
We verify three properties according to Theorem 4.3. First, W is nonempty because the origin 0 ' W . Ifu, v 'W , then
u =
!
"xy0
#
$ and v =
!
"x"
y"
0
#
$ =" u + v =
!
"x + x"
y + y"
0
#
$ 'W.
Hence W is closed under addition. Similarly, if c is any real number and u 'W , then
u =
!
"xy0
#
$ =" cu = u =
!
"c xc y0
#
$ 'W.
Hence W is closed under scalar multiplication. This shows that W is indeed a subspace.
MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS 9
#18. Page 206; Exercise 4. Consider the unit square shown in the accompanying figure. Let W be the set
of all vectors of the form(
xy
), where 0 ) x ) 1, 0 ) y ) 1. That is, W is the set of all vectors whose tail
is at the origin and whose head is a point inside or on the square. Is W a subspace of R2? Explain.
Solution: No, W is not a subspace of R2. W is not closed under either addition or scalar multiplication,because if we choose
u = v =(
11
)'W =" 2u = u + v =
(22
)('W.
#19. Page 206; Exercise 6. In Exercises 5 and 6, which of the given subsets of R3 are subspaces? The setof vectors of the form
(a)
!
"ab0
#
$
(b)
!
"abc
#
$, where a > 0.
(c)
!
"aac
#
$
(b)
!
"abc
#
$, where 2 a! b + c = 1.
Solution: (a) This is a subspace. This follows from Exercise 2 (Problem #17 above).
(b) This is not a subspace. We define
W =
01
2x =
!
"abc
#
$ ' R3
%%%% a > 0 ' R
34
5 .
If W were a subspace, then it would be closed under scalar multiplication. In particular 0u = 0 would bean element. However, !
"abc
#
$ =
!
"000
#
$ 0 = a ) 0.
This means 0 ('W . Hence W cannot be a subspace.(c) This is a subspace. We define
W =
01
2x =
!
"aac
#
$ ' R3
%%%% a, c ' R
34
5 .
We verify three properties according to Theorem 4.3. First, W is nonempty because the origin 0 ' W ; thiscan be seen by choosing a = c = 0. If u, v 'W , then
u =
!
"a1
a1
c1
#
$ and v =
!
"a2
a2
c2
#
$ =" u + v =
!
"aac
#
$ in terms of6
a = a1 + a2
c = c1 + c2
10 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
Similarly, if r is any real number and u 'W , then
u =
!
"a1
a1
c1
#
$ =" r u + v =
!
"aac
#
$ in terms of6
a = r a1
c = r c1
(d) This is not a subspace. We define
W =
01
2x =
!
"abc
#
$ ' R3
%%%% 2 a! b + c = 1 ' R
34
5 .
If W were a subspace, then it would be closed under scalar multiplication. In particular 0u = 0 would bean element. However, !
"abc
#
$ =
!
"000
#
$ 0 = a! 2 b + c (= 1.
This means 0 ('W . Hence W cannot be a subspace.
#20. Page 206; Exercise 10. In Exercises 9 and 10, which of the given subsets of the vector space, M23, or2$ 3 matrices are subspaces? The set of all matrices of the form
(a)(
a b cd e f
), where a = 2 c + 1.
(b)(
0 1 ab c 0
)
(c)(
a b cd e f
), where a + c = 0 and b + d + f = 0
Solution: (a) This is not a subspace. If W were a subspace, then it would be closed under scalar multipli-cation. In particular 0u = 0 would be an element. However,
(a b cd e f
)=
(0 0 00 0 0
)=" 0 = a (= 2 c + 1 = 1.
This means 0 ('W . Hence W cannot be a subspace.(b) This is not a subspace. The 2$ 3 zero matrix is not in the space because
(0 0 00 0 0
)(=
(0 1 ab c 0
)
for any real numbers a, b, and c.(c) This is a subspace. We define
W =6(
a b cd e f
)'M23
%%%% a + c = 0, b + d + f = 07
.
We verify three properties according to Theorem 4.3. First, W is nonempty because the origin 0 ' W ; thiscan be seen by choosing a = b = c = d = e = f = 0. If u, v 'W , then
