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Math 107 August 6, 2009 (Partial) Solution Manual

107 (Partial) Solution Manual

1 Matrices and Determinants

1.1 Systems of Linear Equations

2. Solve the system:

2x +y 2z = 02x y 2z = 0

x +2y 4z = 0

2 1 2 02 1 2 01 2 4 0

rref

1 0 0 0

0 1 0 0

0 0 1 0

x = 0, y = 0, z = 0


x1 +x2 x3 +2x4 = 1x1 +x2 x3 x4 = 1x1 +2x2 +x3 +2x4 = 1

2x1 +2x2 +x3 +x4 = 2

1 1 1 2 11 1 1 1 11 2 1 2 12 2 1 1 2

rref

1 0 0 0 113

0 1 0 0 1030 0 1 0 23

0 0 0 1 23

x1 =113 , x2 = 103 , x3 = 23 , x4 = 23


2x1

x2

x3 +x4 +x5 = 0

x1 x2 +x3 +2x4 3x5 = 03x1 2x2 x3 x4 +2x5 = 0

2

1

1 1 1 0

1 1 1 2 3 03 2 1 1 2 0

rref 1 0 0 7

4 0

0 1 0 9 5 00 0 1 4 4 0

x4 and x5 are free variables, while x1 = 7x4 + 4x5, x2 = 9x4 + 5x5, x3 = 4x4 + 4x5

17. Determine conditions on a, b, and c, so that system is consistent:

2x y +3z = ax 3y +2z = bx +2y +z = c

2 1 3 a1 3 2 b1 2 1 c

2 1 3 a0 52 12 b 12a0 0 0 c + b a

Thus, c + b a = 0 for the system to be consistent.19. Determine conditions on a, b, c, and d, so that system is consistent:

1 1 1 1 a1 1 1 1 b1 1 1 1 c

1 1 1 1 d

rref

1 0 0 0 12(a + b)

0 1 0 0 12(c d)0 0 1 0 12(d b)0 0 0 1 12(c a)

Thus, there are no conditions on a, b, c, and d for the system to be consistent.

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22. In this case, we have 4 unknowns and 3 equations in a homogeneous linear system. By

Theorem 1.1, there are infinitely many solutions.

23. Are there any nontrivial solutions to the system?

x y +z = 02x +y +2z = 0

3x 5y +3z = 0 1 1 1 02 1 2 0

3 5 3 0

1 1 1 00 3 0 0

0 0 0 0

Yes, since in the process of row reduction, we get a row of zeroes.

26. There are infinitely many answers to each part of this problem. Here are some examples:

One Solution:x +2y = 1

x

y = 0

No Solution:x +2y +3z = 0

x +2y 3z = 0x 2y z = 1

Infinitely Many Solutions:x +2y +3z = 0

x +2y 3z = 02x +4y z = 0

28. There are 4 possibilities. With three variables, each equation represents a plane in 3-D space.There are 4 ways that three planes can intersect: (1) No intersection, (2) Intersection at a

single point, (3) Intersection on a line, and (4) All three planes are identical (intersection

everywhere on that plane).

1.2 Matrices and Matrix Operations

5.

A 4B =

7 615 7

2 17

9.

EF =

0 8 93 5 133 4 10

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11. AE is not a valid multiplication; it is undefined.

12.

EA =

2 0

3 4

4 1

14.

B(C + D) =

0 414 816 16

18. A3 = A A A. A A is not defined, since A is not a square matrix.

20.

1

3 1

5

1 1 1 11 1 1 6

x1

x2

x3

x4

=

2

1

6

21.2x1 2x2 +5x3 +7x4 = 124x1 +5x2 11x3 +3x4 = 3

23. A and B are n n matrices:

(a) (A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B) = A2 + AB + BA + B2

(b) In general, AB = BA. This implies that AB + BA = 2AB.

28. There are many possible answers to this question. Here is one:

A =

1 0

0 0

, B =

0 0

0 1

, AB =

0 0

0 0

29. (a) A = [ a1 a2 an ], B =

B1

B2

B3

...

Bn

. Now,

AB = A

B1

B2...

Bn

= A

B1

0

0...

0

+

0

B2

0...

0

+ . . . +

0

0...

0

Bn

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= [ a1 a2 an ]

B1

0

0...

0

+ [ a1 a2 an ]

0

B2

0...

0

+ . . . [ a1 a2 an ]

0

0...

0

Bn

= AB = a1B1 + a2B2 + . . . + anBn

(b)

AB = 2

12

4

+

1

1

1

+ 6

0

1

2

=

3

3

15

30. (a) B =

b1

b2...

bn

, A = [ A1 A2 An ]. Now,

AB = [ A1 A2 An ]

b1

b2...

bn

= ([ A1 0 0 ] + [ 0 A2 0 0 ] + . . . + [ 0 0 An ])

b1

b2...

bn

= [ A1 0 0 ]

b1

b2...

bn

+ [ 0 A2 0 0 ]

b1

b2...

bn

+ . . . + [ 0 0 An ]

b1

b2...

bn

= AB = b1A1 + b2A2 + . . . + bnAn

(b)

AB = 2

3

0

1

2

3

1

+ 3

15

2

=

1

12

7

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1.3 Inverses of Matrices

1. Inverse:

17 27

37 17

6. Inverse:

12 0 12

111 311 0

411 111 0

7. Inverse:

11

12

9

12

1

12 1

4

14 14 14 14

1 12 12 0

712 14 112 14

10. Inverse times right hand side:

11

12

9

12

1

12 1

4

14 14 14 14

1 12 12 0

712 14 112 14

3

5

1

2

=

73

12

74

0

2912

x1 =

7312 , x2 = 74 , x3 = 0, and x4 = 2912 .

11b. Replace the first row by itself times twice the second row: E =

1 2

0 1

.

13. Proof of Theorem 1.6: Let E = [eij ] and A = [aij ], so that EA = [n

k=1

eikakj ].

(a) Suppose E switches rows l and m of I. This means that eij = 1 if i equals j and they

are not equal to l or m or if i = m and j = l or if i = l and j = m. Otherwise eij = 0.

Consider the ij entry of EA:

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If i is not equal to l or m, then the only nonzero eik is eii, so

entij(EA) =n

k=1

eikakj = eiiaij = aij .

Therefore, the ith

row of EA is the ith

row of A If i = l, then the only nonzero elk is elm, so

entlj(EA) =n

k=1

elkakj = elmamj = amj .

Therefore, the lth row of EA is the mth row of A.

If i = m, then the only nonzero emk is eml, so

entmj(EA) =n

k=1

emkakj = emlalj = alj .

Therefore, the mth row of EA is the lth row of A, leaving the rest of A the same.

Thus, EA switches rows l and m of A.

(b) Suppose E multiplies row l of I by c. Then, ell = c, and eii = 1 for all i = l and eij = 0for i = j. Consider the ij entry of EA:

If i = l, then the only nonzero eij is eii = 1,

entij(EA) =n

k=1


Therefore, the ith row of EA is the ith row of A.

If i = l, then the only nonzero elj is ell = c,

entlj(EA) =n

k=1

elkakj = ellalj = calj .

Therefore, the lth row of EA is the lth row of A times c.

Thus, EA multiplies the lth row of A by c, leaving the rest of A the same.

(c) Suppose E replaces row l by itself plus c times row m. Then, eij = 1 if i = j, elm = c,

and eij = 0 for i = j for all other i and j. Consider the ij entry of EA: If i = l, then the only nonzero eij is eii = 1,

entij(EA) =n

k=1


Therefore, the ith row of EA is the ith row of A.

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17. A + B is not symmetric:

A + B =

4 2 21 4 62 6 6

20. BTB is symmetric for any square matrix B (Theorem 1.14, Part 3). In this case,

BTB =

11 2 22 20 102 10 6

22c. Let A = [aij ] be an n n upper triangular matrix. (The proof for a lower triangular matrixis similar.) () Assume A has all diagonal entries nonzero. This implies that we can do rowoperations on A (multiply row i by 1

aii) to get an upper triangular matrix with ones on the

diagonal. These ones can then be use to eliminate everything above in the column using rowoperations (Eliminate off-diagonal aij by replacing row i by row i plus aij times row j, etc.).This leaves the identity. Thus, A is invertible.

( = ) (Contrapositive) Assume one of the diagonal entries is zero. If ann = 0, then the lastrow is zero, and the reduced row echelon form for A will have a zero row, and, thus, A will

not be invertible. Otherwise, let m be the largest m such that amm = 0, i.e. ifi > m, aii = 0.Similar to above, we can do row operations on A (multiply row i by 1

aii) for all rows below

row m, to get an upper triangular matrix with ones on the diagonal for all rows beyond row

m. These ones can then be use to eliminate everything in row m by successively replacing

row m by row m plus amj times row j for j > m. Since amm = 0, row m will then be a zerorow. Thus, the reduced row echelon form for A will be zero, and A will not be invertible.

24d. Since A is an symmetric matrix, A = AT. Since A is invertible, it is also true that

A1 = (AT)1

By Theorem 1.13, Part(5), for any square matrix, (AT)1 = (A1)T. Thus,

A1 = A1)T

Thus, A1 is symmetric.

26. ( = and ) A and B are row equivalent if and only if there exists elementary matrices,{Ek} such that

EkEk1 E1A = B

Let C = EkEk1 E1. By Theorem 1.10, C is the product of elementary matrices if andonly if C is invertible. Thus, C is invertible.

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32. A3 =

0 0 0

0 0 0

0 0 0

.

33. Assume A = [aij ] is an n

n upper triangulr matrix with zeros along the diagonal. (The

proof is similar for a lower triangular matrix.) The proof will proceed by induction on the

number of superdiagonals that are zero, where the kth superdiagonal is define by the entries

{ai(i+k)} for i = 1 . . . n k, the entries whose columns are k over from the diagonal.

Base Case k = 1:

A2 =

0 a12 a13 a1n0 0 a23 a2n...

.... . .

...

0 0 0 a(n1)n0 0 0 0

0 a12 a13 a1n0 0 a23 a2n...

.... . .

...

0 0 0 a(n1)n0 0 0 0

=

0 0 b13 b1n0 0 0 b24 b2n...

.... . .

. . ....

0 0 0 b(n2)n0 0 0 00 0

0 0

where bij are some nonzero numbers. Note that every entry 1 column over from the

diagonal is 0.

Induction Step. Assume

Ak =

0 0 b1(k+1) b1n...

.... . .

. . ....

0 0 0 b(nk)n0

0 0 0 0

......

. . .. . .

...

0 0 0 0 0

where every entry in the kth, (k 1)th, . . . , and 1st superdiagonals are zero, i.e every

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of a 1 1 lower triangular matrix is just the single entry. Assume that the determinant of alower triangular n n matrix is the product of the diagonals. Let A be a (n + 1) (n + 1)lower triangular matrix,

A =

a11 0 0

a21 a22 0 0... ... . . . ...an1 an2 ann

Lets do the cofactor expansion of A about the first row to find the determinant. This gives:

det(A) =

a11 0 0a21 a22 0 0

......

. . ....

an1 an2 ann

= a11

a22 0 0a32 a33 0 0

......

. . ....

an2 an3 ann

= a11det(B1)

Now, the matrix B1 is an n n lower triangular matrix, so the determinant of B is theproduct of the diagonals. Thus,

det(B1) = a22 ann

and

det(A) = a11a22 ann

16. Let A be a square matrix such that row i is c times row j. Let B the matrix obtained from A

by replacing row i by itself plus c times row j. Thus, row i of B is a zero row. By Corollary1.17, det(B) = 0. By Theorem 1.20, det(B) = det(A). Thus, det(A) = 0.

If a square matrix has a column that is a scalar multiple of another column, then its deter-

minant is also zero. In this case, take the transpose of the matrix and apply the just proven

statement. Then, by Theorem 1.19, the transpose and the original matrix have the same

determinant.

1.6 Further Properties of Determinants

4. The determinant of the matrix is 0. The matrix is not invertible.

6. A =

2 31 2

= det(A) = 1.

adj(A) =

2 31 2

= inv(A) = adj(A)

det(A)=

2 31 2

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10.

A =

5 4 12 3 23 1 3

and det(A) = 24

A1 =

2 4 14 3 22 1 3

and det(A1) = 60

A2 =

5 2 1

2 4 2

3 2 3

and det(A2) = 48

A3 =

5 4 22

3 4

3 1 2

and det(A3) = 60

Thus, x = 2.5, y = 2, and z = 2.5.

13. Suppose E is an elementary matrix.

(a) If E is obtained from I by multiplying a row of I by a scalar c, then by Theorem 1.20,

part (2), det(E) = cdet(I) = c.

(b) If E is obtained from I by replacing a row of I by itself plus c times another row, then

by Theorem 1.20, part (3), det(E) = det(I) = 1.

15c. det(A + B) = 28, det(A) = 14, det(B) = 7, det(A) + det(B) = 21. Thus, det(A + B) =det(A) + det(B).

16 If A and B are square matrices of the same size, then, by Theorem 1.24, det(AB) =

det(A)det(B) and det(BA) = det(B)det(A). But, determinants of matrices are real num-

bers, so det(A)det(B) = det(B)det(A). Thus, det(AB) = det(BA).

1.7 Proofs of Theorems on Determinants

5. Suppose A is an n n matrix. Consider the ij entry of adj(A)A:

entij(adj(A)A) =n

k=1

Cikakj

if i = j, then

entii(adj(A)A) =n

k=1

akiCik = det(A)

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since the sum is the cofactor expansion of A about column i. If i = j, then

entij(adj(A)A) =n

k=1

akjCik

is the determinant of the matrix, call it B, obtained from A by replacing the jth

columnof A by the ith column of A. Since this matrix contains two columns with same entries,

the transpose of B has two identical rows. This implies det(BT) = 0. By Theorem 1.19,

det(B) = 0. Thus, for i = j,entij(adj(A)A) = 0

= adj(A)A = det(A)I

6. Let A =

A1

A2

A3...

An

, where Ai is the ith row of A. This implies cA =

cA1

cA2

cA3...

cAn

.

By application of Theorem 1.20 n times,

det(cA) = cdet

A1

cA2

cA3...

cAn

= c2

det

A1

A2

cA3

...

cAn

= . . . = cn

det

A1

A2

...

An

= c

n

det(A)

7. The following lemma will be helpful.

Lemma 1. If the matrix B is obtained from the matrix A by multiplying one of the columns

of A by a scalar c, then det(B) = c det(A).

Suppose B is obtained from A by multiplying one of the columns of A by a c. Note thatBT can be obtained from AT by multiplying a row of AT by c. By Theorem 1.6, part(2),

det(BT) = c det(AT). By Theorem 1.19, det(AT) = det(A) and det(BT) = det(B). Thus,det(B) = c det(A).

Let Vn be the n n Vandermonde matrix

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Thus,

det(Vn+1) =

1 1 1 10 x2 x1 x3 x1 xn+1 x10 x22 x21 x23 x21 x2n+1 x21... ... ... ...

0 xn2 xn1 xn3 xn1 xnn+1 xn1

=

x2 x1 x3 x1 xn+1 x1x22 x21 x23 x21 x2n+1 x21

..

.

..

.

..

.xn2 xn1 xn3 xn1 xnn+1 xn1

By Lemma 1 and the fact that for any positive integer m,

ym+1 zm+1 = (y z)

mk=0

ymkzk

the determinant becomes

det(Vn+1) = (x2x1)(x3x1) (xn+1x1)

1 1 1x2 + x1 x3 + x1

xn+1 + x1

... ... ...n2k=0

xn2k2 xk1

n2k=0

xn2k3 xk1

n2k=0

xn2kn+1 xk1

n1k=0

xn1k2 xk1

n1k=0

xn1k3 xk1

n1k=0

xn1kn+1 xk1

where the second to last row is now explicitly shown. Consider row i of the matrix

above:

i1

k=0

xi1k2 xk1

i1

k=0

xi1k3 xk1

i1

k=0

xi1kn+1 xk1

If we multiply row i by x1, it becomes

ik=1

xik2 xk1

i1k=1

xik3 xk1

n1k=1

xikn+1xk1

Thus, adding x1 times row i to row i + 1,i

k=0

xik2 xk1

ik=1

xik2 xk1

ik=0

xik3 xk1

i1k=1

xik3 xk1

ik=0

xikn+1xk1

n1k=1

xikn+1xk1

= xi2 xi3 xin+1 By Theorem 1.20, adding a multiple of one row to another does not change the deter-

minant, so if for each i we add x1 times row i to row i + 1, we obtain

det(Vn+1) = (x2 x1)(x3 x1) (xn+1 x1)

1 1 1x2 x3 xn+1...

......

xn22 xn23 xn2n+1

xn12 xn13 xn1n+1

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2. (b) Not a subspace since it is not closed under scalar multiplication:

c

y + z + 1

y

z

=

cy + cz + c

cy

cz

,

which is not in the subset if c = 1.(d) Not a subspace since it is not closed under scalar multiplication:

c

x

y

x2 + y2

=

cx

cy

cx2 + cy2

,

which is not in the subset for all c.

3. (a) Subspace

(b) Not a subspace (not closed under both addition and scalar multiplication)(c) Subspace

(d) Subspace

(e) Not a subspace (not closed under both addition and scalar multiplication)

5. The solutions to that linear system do not define subspace. The set of solutions is not closed

under addition or scalar multiplication: Let X1 and X2 be two solutions to AX = B, with B

nonzero. Then,

A(X1 + X2) = AX1 + AX2 = B + B = 2B

= B

A(cX1) = cAX1 = cB = B, if c = 1

11. Consider 1 2

1 3

c1

c2

=

x1

x2

Since

1 21 3 = 5 = 0, given x1 and x2, this system has a unique solution for c1 and c2.

Thus, it spans R2. Since the solution c1 = c2 = 0 is unique for x1 = 0 and x2 = 0, the vectors

are also linearly independent. Thus, they are a basis for R2.

12. Consider 1 5

2 4

c1

c2

=

x1

x2

Since

1 52 4 = 6 = 0, given x1 and x2, this system has a unique solution for c1 and c2.

Thus, it spans R2. Since the solution c1 = c2 = 0 is unique for x1 = 0 and x2 = 0, the vectors

are also linearly independent. Thus, they are a basis for R2.

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The matrix has determinant 1, so the system has a unique solution, c1 = c2 = c3 = c4 = 0.

Thus, the vectors are linearly independent.

14. Assume c1

2

10

+ c2

1

3

1

+ c3

1

41

= 0. Since

det(A) =

2 1 1

1 3 41 3 4

= 14 = 0

there is a unique solution to the coefficient problem, A

c1

c2

c3

= b. If b = 0, then since the

solution is unique,

c1c2c3

= 0, so the vectors are linearly independent. For any other b, asolution exists, so the vectors span R3. Therefore, the vectors form a basis.

17. Assume c1(x2 + x + 1) + c2(x

2 x + 1) + c3(x2 1) = 0. Then,

1 1 1

1 1 01 1 1

c1

c2

c3

= 0

The matrix has determinant equal to 4, so the system has a unique solution c1 = c2 = c3 = 0.

Thus, the vectors are linearly independent. The nonzero determinant also implies a unique

solution to the nonhomogeneous problem, which implies that the vectors span P2. Thus, the

vectors are a basis.

21. Consider c1(x + 1) + c2(x + 2) + c3(x + 3) = 0 or the equivalent system

1 1 1

1 2 3

c1

c2

c3

= 0

Apply row operations, the matrix above becomes1 0 10 1 2

Thus, c1 = c3, and c2 = 2c3 with c3 a free variable are solutions. For example, c1 = 1,c2 = 2, c3 = 1 is a solution. Thus, the vectors are not linearly independent and, therefore,not a basis.

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4. (c) c1

0 1

0 1

+ c2

1 0

0 1

+ c3

1 0

0 0

+ c4

0 1

1 0

= 0 implies

c2 + c3 = 0

c1

+ c4

= 0

c4 = 0

c1 + c2 = 0

= c1 = 0 = c2 = 0 = c3 = 0. Thus, the vector matrices are linearlyindependent. Since there are 4 of them, they form a basis for M22(R).

(d) There are 5 vector matrices in the set and dim(M22(R)) = 4, so the set cannot be a

basis.

7. 1 1 1

1 1 01 1 2

rref

1 1 00 0 1

0 0 0

Thus, x2 is a free variable and x1 x2 = 0, x3 = 0. Thus, every vector in the nullspace is

of the form:

x2

x2

0

. Thus, a basis for the nullspace is {

1

1

0

}. A basis for the rowspace is

{[1 -1 0], [0 0 1]}.

1

1 1

1 1 01 1 2

rcef 1 0 0

0 1 0

2 1 0

So, a basis for the column space is {

1

0

2

,

0

1

1

}. Thus, the rank of the matrix is 2.

10. 2 1 3 41 0

1 3

rref

1 0 1 30 1

5 2

Thus, x3 and x4 are free variables and x1 x3 + 3x4 = 0, x2 5x3 + 2x4 = 0. Therefore, anyvector in the nullspace can be written

x3

5x3

x3

0

+

3x42x4

0

x4

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Therefore, a basis for the nullspace is {

1

5

1

0

,

320

1

}. A basis for the rowspace is {[1 0 -1 3],

[0 1 -5 2]}. 2 1 3 41 0 1 3

rref

1 0 0 0

0 1 0 0

So, a basis for the column space is {

1

0

,

0

1

}. Thus, the rank of the matrix is 2.

14.

0 1 2 4

2 1 0 2

1 0 1 1

rref

1 0 1 00 1 2 0

0 0 0 1

So, the rank of the matrix is 3. Thus, the dimension of the subspace spanned by the vectors

is 3, so it is all ofR3. So, a basis for the subspace is {

1

0

0

,

0

1

0

,

0

0

1

}

18. Suppose A is an m n matrix, with m > n. Let B be the reduced row echelon form forA. Then, the RS(A) = RS(B) by Theorem 2.13, since A and B are row equivalent. Also,

since m > n, B has at most n nonzero rows and since B is in reduced row echelon form, the

nonzero rows of B form a basis for RS(B), and thus, RS(A). Therefore, the set of rows of

A is subset of RS(A) with more vectors in it than the basis. Thus, by Lemma 2.8, the set of

rows is linearly dependent.

2.5 Wronskians

5. w(x2

1, x2 + 1, x + 1) =

x2 1 x2 + 1 x + 12x 2x 1

2 2 0

= 4

= 0 for some x, so the functions are

linearly independent.

6. w(ex, e2x, e3x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

= 2e6x = 0 for some x, so the functions are linearly

independent.

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12. By trigonometric identity, for all x,

cos(2x) = cos2(x) sin2(x)

Thus, for all x,

cos(2x) cos2

(x) + sin2

(x) = 0

Therefore, the functions are linearly dependent.

14. (a) No, since one (or all) of the functions could be zero on the interval (c, d), and linearly

dependent there, but be, for example, x, x2, x3,. . . , etc. on the rest of the interval and

stil be linearly independent on (a, b).

(b) Yes, since if there are no scalars to make a linear combination of the functions equal to

zero on all of (c, d), there cannot be scalars to make a linear combination of the functions

equal to zero on all of any interval (a, b) that includes (c, d).

3 First Order Ordinary Differential Equations

4 Linear Differential Equations

4.1 The Theory of Higher Order Linear Differential Equations

2. 2nd order, linear

3. 4th order, linear

6. (1, )

10. The two functions solve the ODE:

1x

: x :

x2( 2x3

) + x( 1x2

) 1x

= 0 x2(0) + x(1) x = 0

The two functions are linearly independent since

1x

x

1x2

1

= 2x , which is nonzero every-where. Thus, they are a fundamental set of solutions and the general solution is given by

c11

x+ c2x

Solving using the initial conditions, y(1) = 0, y(1) = 1,

c1 + c2 = 0

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c1 + c2 = 1

gives, c1 = 12 , c2 = 12 , which gives the solution 12x + 12x.

11. The three functions solve the ODE:

ex : e2x : e3x :

ex 7ex + 6ex = 0 8e2x 7(2e2x) + 6e2x = 0 27e3x 7(3e3x) + 6e3x = 0

The Wronskian of the three functions is

ex e2x e3x

ex 2e2x 3e3xex 4e2x 9e3x

= 20

so the functions are linearly independent. The general solution to the ODE is

c1ex + c2e

2x + c3e3x

Using the initial condition, y(0) = 1, y(0) = 0, y(0) = 0 gives

c1 + c2 + c3 = 1, c1 + 2c2 3c3 = 0, c1 + 4c2 + 9c3 = 0

which has solution c1 =32 , c2 = 35 , and c3 = 110 , giving the solution to the ODE as

3

2ex

3

5e2x +

1

10e3x

15. The general solution is given by

ex + c1ex + c2e2x + c3e3x

For the initial conditions, y(0) = 1, y(0) = 0, y(0) = 0, this gives

1 + c1 + c2 + c3 = 1, 1 + c1 + 2c2 3c3 = 0, 1 + c1 + 4c2 + 9c3 = 0

which has solution c1 = 0, c2 =1

5

, c3 =

1

5

, giving the solution to the ODE as

ex +1

5e2x 1

5e3x

17. {x, x2} is a fundamental set of solutions:

x : x2 :

12x

2(0) x(1) + x = 0 12x2(2) x(2x) + x2 = 0

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(b) The initial conditions imply that c1 c2 + c3e5 = 1, c2 + 5c3e5 = 0, and 25c3e5 = 2.Thus, c1 =

1325 , c2 =

25 , c3 =

2e5

25 so the solution to the IVP is

13

25 2

5x +

2

25e5(x+1)

23. (a) The characteristic polynomial has roots 0, 2, and 3, the general solution is

c1 + c2e2x + c3e3x

(b) The initial conditions imply that c1 + c2 + c3 = 1, 2c2 + 3c3 = 0, and 4c2 + 9c3 = 2.Thus, c1 =

23 , c2 =

15 , c3 =

215 so the solution to the IVP is

2

3+

1

5e2x +

2

15e3x

24. The hard way.

(a) The characteristic polynomial has roots 0 and 2 i, the general solution is

c1 + c2e2x sin x + c3e

2x cos x

(b) The initial conditions imply that c1+c2e2 sin(1)+c3e

2 cos(1) = 1, e2(2 sin(1)+cos(1))c2+

e2(2 cos(1) sin(1)c3 = 0, and e2(3 sin(1) + 4 cos(1))c2 + e2(4 sin(1) + 3 cos(1))c3 = 2.Thus, c1 =

75 , c2 = e

2(45 cos(1) 25 sin(1)), c3 = e2(25 cos(1)+ 45 sin(1)), so the solutionto the IVP is

7

5+ e2(

4

5cos(1) 2

5sin(1))e2x sin(x) + e2(2

5cos(1) +

4

5sin(1))e2x cos(x)

24. The easy way.

(a) The characteristic polynomial has roots 0 and 2 i, the general solution is

c1 + c2e2x2 sin(x 1) + c3e2x2 cos(x 1)

(b) The initial conditions imply that c1 + c3 = 1, c2 + 2c3 = 0, and 4c2 + 3c3 = 2. Thus,

c1 =75 , c2 =

45 , c3 =

25 , so the solution to the IVP is

7

5+

4

5e2x2 sin(x 1) 2

5e2x2 cos(x 1)

The two solutions can be seen to be identical by the fact that

sin(AB) = sin(A) cos(B) cos(A)sin(B) and cos(AB) = cos(A)cos(B) sin(A) sin(B)

29. The general solution is c1e2x + c2e3x + c3xe3x + c4e3x + c5xe3x.

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30. The general solution is

c1 + c2x + c3x2 + c4e

4x + c5xe4x+

exc6 cos(

2

2x) + c7 sin(

2

2x) + c8x cos(

2

2x) + c9x sin(

2

2x)+

e1

2x

c10 cos(

3

2x) + c11 sin(

3

2x)

40. The conditions on the roots r1, . . . , rn real and distinct, so that the equation decays to 0 as

x are that ri < 0 for all i.

41. The conditions on the roots r1, . . . , rn real, but not necessarily distinct, so that the equation

decays to 0 as x are that ri < 0 for all i.

42. The conditions on the roots r1, . . . , rn possibly complex, so that the equation decays to 0 as

x are that Re(ri) < 0 (the real part is negative) for all i.

4.3 The Method of Undetermined Coefficients

1. The characteristic polynomial of the homogeneous equation has roots 3 and 2, so a fun-damental set for the homogeneous equation is e3x, e2x. Trying a particular solution of the

form Ae2x yields

4Ae2x 2Ae2x 6Ae2x = 3e2x

or A = 34 . Thus, the general solution is

34

e2x + c1e3x + c2e

2x

4. The characteristic polynomial of the homogeneous equation has roots 1 and 12 , so a fun-damental set for the homogeneous equation is ex, e

1

2x. Trying a particular solution of the

form A cos(x) + B sin(x) yields

2A cos(x) 2B sin(x) 3A sin(x) + 3B cos(x) + A cos(x) + B sin(x) = cos(x)

or A + 3B = 1 and B 3A = 0, which has solution A = 110 , B = 310 . Thus, the generalsolution is

110

cos(x) +3

10sin(x) + c1e

x + c2e1

2x

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9. The characteristic polynomial of the homogeneous equation has roots 2i, so a fundamentalset for the homogeneous equation is cos(2x), sin(2x). Trying a particular solution of the form

Ax cos(2x) + Bx sin(2x) yields

16A sin(2x)+16B cos(2x)

16Ax cos(2x)

16Bx sin(2x)+16Ax cos(2x)+16Bx sin(2x) = 3cos(2x)

or 16A = 0 and 16B = 3, which has solution A = 0, B = 316 . Thus, the general solution is3

16x sin(2x) + c1 cos(2x) + c2 sin(2x)

11. The characteristic polynomial of the homogeneous equation has roots 0 and 8, so a funda-mental set for the homogeneous equation is 1, e8x. Trying a particular solution of the form

Ax3 + Bx2 + Cx yields

6Ax + 2B + 24Ax2

+ 16Bx + 8C = 2x2

7x + 3or 24A = 2, 6A + 16B = 7 and 2B + 8C = 3, which has solution A = 112 , B = 1532 , C = 63128 .Thus, the general solution is

1

12x3 15

32x2 +

63

128x + c1 + c2e

8x

14. The characteristic polynomial of the homogeneous equation has roots 2 and 4, so a fun-damental set for the homogeneous equation is e2x, e4x. Trying particular solutions of the

forms Ax2 + Bx + C, Dxe2x and F cos(4x) + G sin(4x) yields

6A + 12Ax + 6B 24Ax2 24Bx 24C = 3x2

12Dxe2x + 12De2x + 6De2x + 12Dxe2x 24Dxe2x = 5e2x

48F cos(4x)48G sin(4x)24F sin(4x)+24G cos(4x)24F cos(4x)24G sin(4x) = 6 sin(4x)

or 24A = 3, 12A24B = 0, 6A+6B24C = 0, 18D = 5, 72F+24G = 0, 72G24F =6, which has solution A = 18 , B = 116 , C = 364 , D = 518 , F = 140 , G = 340 . Thus, thegeneral solution is

18

x2 116

x 364

518

xe2x +1

40cos(4x) +

3

40sin(4x) + c1e

2x + c2e4x

18. The characteristic polynomial of the homogeneous equation has a double root 2, so a fun-damental set for the homogeneous equation is e2x, xe2x. Trying particular solutions of the

forms Ax + B and Ccos(3x) + D sin(3x) yields

4A + 4Ax + 4B = 2x

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So,

yp(x) =1

2

3x2ex 6xex 5ex ex 12

3x2ex 6xex + 5ex ex

= 3

2 x2

3x 5

2 3

2 x2

+ 3x 5

2 = 3x2

5

2. The roots of the characteristic polynomial are 3 and 2, making e3x and e2x a fundamentalset of solutions to the homogeneous equation. w(e3x, e2x) = 5ex, so, then, yp = u1(x)e3x +u2(x)e

2x, with

u1(x) =

e2x4ex

5ex = 4

10e2x

and

u2(x) =e3x4ex5ex = 415 e3x

so,

yp(x) = 410

e2xe3x 415

e3xe2x

= 23

ex

So, the general solution is

y = 2

3 ex

+ c1e3x

+ c2e2x

3. The characteristic polynomial has a double root at 3, so a fundamental set of solutions ise3x, xe3x. Thus, w(e3x, xe3x) = e6x. Then, if yp(x) = u1(x)e3x + u2(x)xe3x,

u1(x) =

3e3xxe3x

e6x= 1

2xe6x +

1

12e6x

and

u2(x) =

3e3xe3x

e6x=

1

2e6x

Thus,

yp(x) = (12

xe6x +1

12e6x)e3x +

1

2e6xxe3x

=1

12e3x

So, the general solution is

y =1

12e3x + c1e

3x + c2xe3x

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5 Linear Transformations and Eigenvalues and Eigenvectors

5.1 Linear Transformations

3. Not a linear transformation: T(x1 + x2, y1 + y2, z1 + z2) = T(x1, y1, z1) + T(x2, y2, z2).4. Linear transformation

7. Linear transformation

12. Not a linear transformation: T(A + B) = det(A + B) = det(A) + det(B) = T(A) + T(B).

18.

1 1 3 12 3

1

2

3 7 5 3

20. Let

A =

1 1 0

1 0 10 1 1

Then,

A1 =

1 1 10 1 1

1 0 1

(a) A1

2

1

4

=

5

36

. So, T

2

1

4

= 5T

1

1

0

3T

10

1

+ 6T

0

1

1

or

T

2

1

4

=

5

0

50

+

630

0

+

6

0

0

6

=

5

35

6

(b) A1

x

y

z

=

x y zy + z

x z

. So,

T

x

y

z

= (x y z)T

1

1

0

+ (y + z)T

10

1

+ (x z)T

0

1

1

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or

T

x

y

z

=

x y z0

x + y + z

0

+

2y + 2z

y + z

0

0

+

x z0

0

x

+z

=

2x + y

y + z

x + y + z

x

+z

33. Suppose

x

y

=

r cos

r sin

. Then,

T

x

y

=

r cos( + )

r sin( + )

=

r cos cos r sin sin r sin cos + r cos sin

=

x cos y sin y cos + x sin

=

cos sin sin cos

xy

35. For any linear transformation, T : V V, either dim(ker(T)) = 0 or dim(ker(T)) > 0, butnot both can be true. The second inequality is 2. We show that dim(ker(T)) = 0 implies 1.

Let v1, . . . , vn be a basis of the n-dimensional vector space V. Then, T(v1), . . . , T(vn) are a

set of n vectors in V. Assume there are scalars {ci} so that

c1T(v1) + c2T(v2) + . . . + cnT(vn) = 0

Then,

T(c1v1 + c2v2 + . . . + cnvn) = 0

Since dim(ker(T)) = 0, this implies that c1v1 + . . . + cnvn = 0, and consquently, that c1 =

c2 = = cn = 0, since v1, . . . , vn is a basis. Thus, T(v1), . . . , T(vn) are linearly independentvectors in V. Since there are n of them, they also form a basis for V. Thus, they also span

V. Therefore, for any vector v V, there exist {ai} so that

v = a1T(v1) + a2T(v2) + . . . + anT(vn) = T(a1v1 + . . . + anvn)

Defining u = a1v1 + . . . + anvn, gives T(u) = v and statement 1.

36. If T : V W is 1-1, then T(x1) = T(x2) = x1 = x2. Suppose u ker(T). Then,T(u) = T(0) = u = 0. So, the kernel is only the zero vector.Ifker(T) is only the zero vector, suppose T(x1) = T(x2) for x1, x2 V. Then, T(x1x2) = 0,which implies x1x2 ker(T). But, zero is the only vector in the kernel ofT, so x1x2 = 0,or x1 = x2. So, T is 1-1.

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5.2 The Algebra of Linear Transformations

5. ST

x

y

= S

x + 3y

x y

=

x + 7y

3x + y

6. T S

xy

= T

2x yx + 2y

=

5x + 5yx 3y

8. T S(ax + b) = T(ax 2a + b) = ax + 2a 2a + b = ax + b

11. Basis={e3x, ex}

14. Basis={sin x, cos x, x sin x, x cos x}

17. Let f(x), h(x) C. Then,

Tg(x)(f + h) = g(x)(f(x) + h(x)) = g(x)f(x) + g(x)h(x) = Tg(x)(f) + Tg(x)(h)

and

Tg(x)(cf) = g(x)(cf(x)) = cg(x)f(x) = cTg(x)

So, Tg(x) is a linear transformation.

20. (a) Ifu ker(T), then T(u) = 0. T(u + v) = T(u) + T(v) = T(v).(b) If T(u) = T(v), then T(u) T(v) = 0 = T(u v) = 0, so that u v ker(T).

23. (a) A =1 3

1 1

, B =2 11 2

(b) C =

1 7

3 1

, BA =

2 11 2

1 3

1 1

=

1 7

3 1

5.3 Matrices for Linear Transformations

1. (a) [T] =

1 1

1 1

(b) The change of basis from to is given by P = 1 21 1 .

(c) The change of basis from to is given by P1 =

1 21 1

.

(d) [T] = P1[T]P =

4 72 4

.

4. (Assuming a spurious x2 in the fourth row.)

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(b) The change of basis from to is given by P =

1 1

1 1

.

(c) The change of basis from to is given by P1 =

12

12

12 12

.

(d) [T] = P1[T]P =

0 1

1 0

.

(e) [v] =

23

. [v] = P

1[v] =

1252

.

(f) [T(v)] = [T][v] =

52

12

.

(g) T(v) = 52(sin(x) + cos(x)) 12(sin(x) cos(x)) = 2 sin(x) + 3 cos(x).

9. (a) [T] =

1 0

1

1 1 00 1 1

(b) [T(v)] = [T]

[v] =

1 0 1

1 1 00 1 1

1

23

=

235

(c) T(v) = 2v1 3v2 + 5v3.

5.4 Eigenvalues and Eigenvectors

5. The eigenvalues are 2

3 and 23, with basis eigenvectors

32

1

and

32

1

, respectively.

8. The eigenvalues are 2 and 1, with basis eigenvectors

111

for 2 and

1

0

0

for 1.

9. The eigenvalue is 4 with basis eigenvector

0

01

.

16. The eigenvalues are 2i and 2i with basis eigenvectors

1 2i1

and

1 + 2i

1

, respectively.

17. The eigenvalues are 1, i, and i with basis eigenvectors

1

0

0

,

0

1

i

, and

0

1

i

, respectively.

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20. IfD is diagonal matrix, the eigenvalues of D are the diagonal entries of D. The bases for the

eigenspaces are just the standard basis vectors for Rn.

26. If A is a square matrix, then eigenvalues of A are the so that det(I A) = 0. Sincedet(B) = det(BT) for any square matrix B, det(I

A) = det (I A)

T = det(I AT).

Thus, det(I A) = 0 if and only if det(I AT), so A and AT have the same eigenvalues.

5.5 Similar Matrices, Diagonalization, and Jordan Form

5. A is diagonalizable:

D =

12 0

0 12

, P =

12

3 12

3

1 1

8. A is not diagonalizable, not enough eigenvectors.

9. A is not diagonalizable, not enough eigenvectors.


D =

2i 0

0 2i

, P =

1 2i 1 + 2i

1 1


D =

1 0 0

0 2i 0

0 0

2i

, P =

1 0 0

0 1 1

0 i

i

31. Suppose A and B are similar matrices. Then, A = P1BP for some invertible matrix P.

Thus, A I = P1BP I for any scalar . Therefore,

det(A I) = det(P1BP I)= det

P1(BP P I) = det P1(B PIP1)P

= det(P1)det(B I)det(P)= det(B I)

Thus, A and B have the same characteristic polynomial.

36. Let A be a n n matrix with n distinct eigenvalues. Since the bases of eigenspaces corre-sponding to different eigenvalues are linearly independent and A has n eigenspaces, there are

n linearly independent eigenvectors of A. Thus, A is diagonalizable by Theorem 5.18.

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6 Systems of Linear Differential Equations

6.1 The Theory of Systems of Linear Differential Equations

1. Y = 2c1e

2x + 3c2e3x

4c1e2x

+ 3c2e3x

4 12 1

Y =

4c1e

2x + 4c2e3x 2c1e2x c2e3x

2c1e2x + 2c2e

3x + 2c1e3x + c2e

3x

= Y

4. W(

e2x

0

0

,

0

3 cos(5x)

3 sin(5x)

,

0

sin5x

cos5x

) = e2x(3 cos2(5x) + 3 sin2(5x)) = 3e2x = 0 for all x.

Therefore, the functions are linearly independent on the whole interval (, ).

5. The general solution to the problem is c1ex

c2e2x

. A matrix of fundamental solutions isex 0

0 e2x

.

9. Y(0) =

2

1

gives the specific solution

2ex

e2x

.

17. Being a matrix, M, of fundamental solutions implies that each column is a solution to the

system and the columns are linearly independent. Since det(M) = ed1+d2+...+dnx = 0 for all

x, the columns are linearly independent. Consider the ith column of M: Yi =

0...

edix

...

0

. Then,

Yi =

0...

diedix

..

.0

and DYi =

0...

diedix

..

.0

, so Yi = DYi and Yi is a solution. Since this is true for

all i, M is a matrix of fundamental solutions.

27. (AB) =

4e2x + 8e2x 2ex 2xex 8x 4e4x

2ex + 4x 2ex 2xex + e3x + 3xe3x

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6.2 Homogeneous Systems with Constant Coefficients - Diagonalizable

3. From before, the eigenvalues are 2

3 and 23, with basis eigenvectors

32

1

and

32

1

,

respectively. Therefore, the general solution to Y = AY is

c1

32 e

23x

e23x

+ c2

32 e

3

2

e3

2

=

c132 e23x c232 e

3

2

c1e23x + c2e

3

2

5. The eigenvalues are 1, 2, and 3, with corresponding basis eigenvectors

0

1

0

,

1

22

,

1

11

,


c1

0

ex

0

+ c2 e2x

2e2x2e2x

+ c3 e3x

e3xe3x

= c2e

2x + c3e3x

c1ex 2c2e2x c3e3x2c2e2x c3e3x

11. The eigenvalues are 1, i, and i with basis eigenvectors

1

0

0

,

0

1

i

, and

0

1

i

, respectively.

Therefore the general solution to Y = AY is

c1

ex

0

0

+ c2 0

cos(x)

sin(x) + c3

0

sin(x)

cos(x)

= c1e

x

c2 cos(x) + c3 sin(x)

c2 sin(x) + c3 cos(x)

15. Y(0) =

0

1

1

=

c2 + c3

c1 2c2 c32c2 c3

= c1 = 2, c2 = 1, and c3 = 1. So, the solution is

Y(x) =

e2x e3x2ex

2e2x + e3x

2e2x + e3x

22. The eigenvalues are solutions to 0 = 216 9 = 2 25 = ( 5)( +5), so the eigenvalues

are 5. The corresponding basis eigenvectors are

3

1

and

1

3

. Therefore the general

solution is

c1

3e5x

e5x

+ c2

e5x

3e5x

=

3c1e

5x + c2e5x

c1e5x 3c2e5x

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25. The eigenvalues are 13i with corresponding basis eigenvectors

3i

1

and

3i

1

. The general

solution is then

c1 3ex sin(3x)

ex

cos(x)+ c2

3ex cos(3x)

ex

sin(3x) =

3c1ex sin(3x) + 3c2ex cos(3x)

c1ex

cos(3x) + c2ex

sin(3x)

28. Y =

2 1 0

3 4 0

5 6 3

Y. The eigenvalues are 1, 3 and 5 with basis eigenvectors

22

11

,

0

0

1

, and

2

6

13

. Thus, the general solution is

2c1ex

+ 2c3e5x

2c1ex + 6c3e

5x

11c1ex + c2e

3x 13c3e5x

Thus,

5

3

4

= Y(0) =

2c1 + 2c32c1 + 6c3

11c1 + c2 13c3

= c3 = 1, c1 = 32 , c2 = 672 . So, the solution is 3e

x

+ 2e5x

3ex + 6e5x332 ex + 672 e3x 13e5x

30. The conditions on the eigenvalues are that the real parts of the eigenvalue are strictly negative

(< 0).

6.3 Homogeneous Systems with Constant Coefficients - Nondiagonalizable

17. The matrix has eigenvalue

6 with multiplicity 2 and a single basis eigenvector 11. Using

Schurs Theorem and the Schur Decomposition, we can use Theorem 9.7 to find an orthogonal

matrix P so that PTAP is upper triangular. Using

1

1

as the start, the matrix P found

using Theorem 9.7 is

P =

12 12

12

12

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with PT = P. Thus,U = PTAP =

6 20 6

. The system Z = U Z gives

z1(t) =

6z1(t) + 2z2(t)

z2(t) = 6z2(t)

which has general solution z2(t) = c2e6t, z1(t) = c1e6t + 2c2te6t or c1

e6t

0

+ c2

2te6t

e6t

.

Thus, the general solution to the original problem is P Z or

12

(c1 + c2)e

6t + 2c2te6t

(c1 c2)e6t + 2c2te6t

Using the initial value Y(0) =

10

, we get the system of equations

1

0

=

12

c1 + c2

c1 c2

Thus, c1 = c2 =22 . Therefore, the solution is

x(t)

y(t)

=

e6t + te6t

te6t

6.4 Nonhomogeneous Linear Systems

1. Y =

3 0

8 1

Y +

2

x

. The eigenvalues of the matrix are 3 and 1, with basis eigenvectors

1

2

and

0

1

. Thus, the general solution to the homogeneous linear system is

YH = c1e

3x

2c1

e3x

+c2

ex

and a matrix of fundamental solutions is M =

e3x 0

2e3x ex

. Thus,

M1 =

e3x 0

2ex ex

and YP = M

M1G(x)dx =

e3x 0

2e3x ex

e3x 0

2ex ex

2

x

dx.

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So,

YP =

e3x 0

2e3x ex

2e3x

4ex + xex

dx

= e3x 0

2e3x ex

2

3e3x

4ex + xex ex

=

23

193 + x

Thus, the general solution is

Y = YP + YH =

23 + c1e3x

193 + x + 2c1e3x + c2ex

5. The eigenvalues are 1, 2, and 3, with corresponding basis eigenvectors

0

1

0

,

1

22

,

1

11

,


c1

0

ex

0

+ c2

e2x

2e2x2e2x

+ c3

e3x

e3xe3x

=

c2e2x + c3e

3x

c1ex 2c2e2x c3e3x2c2e2x c3e3x

and a matrix of fundamental solutions is M =

0 e2x e3x

ex 2e2x e3x0 2e2x e3x

. Thus,

M1 =

0 ex exe2x 0 e2x2e3x 0 e3x

and

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Using the initial condition (1, 1) gives the system

1 + c1 + c2 = 1

1 + 2c1 c2 = 1

leading to c1 = 23 , c2 = 23 and the solutionx(t)

y(t)

=

t + 1 + 23e

4t 23ett 1 + 43e4t + 23et

9 Inner Product Spaces

9.1 Inner Product Spaces

6. < v,u >= 0,

||v

||= ,

||u

||= , The angle between u and v is 2 .

7. < f, g >=

ba

f(x)g(x)dx =

ba

g(x)f(x)dx =< g,f >

< f+g,h >=

ba

(f(x)+g(x))h(x)dx =

ba

f(x)h(x)dx+

ba

g(x)h(x)dx =< f,h > + < g,h >

< cf, g >=

ba

cf(x)g(x)dx = c

ba

f(x)g(x)dx = c < f, g >

< f, f >=

ba

f2(x)dx 0 for all f andba

f2(x)dx = 0 ifff(x) = 0 for all x.

8. < v,u >= a1w1b1 + a2w2b2 + . . . + anwnbn =< u, v >

< v + u, w >= (a1 + b1)w1c1 + (a2 + b2)w2c2 + . . . + (an + bn)wncn = a1w1c1 + . . . + anwncn +b1w1c1 + . . . + bnwncn =< v, w > + < u, w >

< cv, u >= ca1w1b1 + ca2w2b2 + . . . + canwnbn = c < v, u >

< v, v >= w1a21 + w2a

22 + . . . + wna

2n 0, since wi > 0 for all i. < v, v >= 0 if and only if

a2i = 0 for all i = ai = 0 for all i = v = 0.

16. Let v be a nonzero vector in a inner product space (V, < , >)). Then ||v|| = < v, v > > 0and

=

< v, v >

||v

||2

= 1

so, v||v|| is a unit vector.

18. 0

sin(x) sin(2x)dx =

0

2sin2(x) cos(x)dx =2

3sin3(x)

0

= 0

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9. < 1, 1 >=

11

(1)(1)dx = 2, so v1 =12

.

w2 = x 1

1

x2

dx

12

= x, v2 =w2||w2|| =

32x.

w3 = x2

1

1

x2

2dx 12

1

1

x3

3

2dxx3

2

= x2

1

3

, v3 =w3

||w3||= 3

5

22(x2

1

3

)

13. Showing the matrix is orthogonal is equivalent to showing that {v1 =

cos

sin

, v2 =

sin

cos

}

is an orthonormal basis for R2.

v1 v1 = cos2 + sin2 = 1,v1 v2 = cos sin + sin cos = 0, andv2 v2 = sin2 + cos2 = 1.Thus, the set is an orthonormal basis and the matrix is orthogonal.

math 107 solutions

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