answers to problem set # 7

8/14/2019 Answers to Problem Set # 7

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EXERCISE 3.1.1

For (a), we apply Laplace expansion along the row or column with the mostnumber of zerosin this case, the third columnto obtain

Da=101010100=-11+3a13M13+-12+3a23M23+-13+3a33M33

=a13C13-a23C23+a33C33

=10110-01010+01001

Da=1.

For (b), we try another useful device for evaluating a third-order determinantthat is, we write down the determinant column by column, then after the third, werepeat the first and second columns to create a 35 array of numbers. The value ofthe determinant is equal to the sum of six terms: the first three terms carry a positivesign and are formed by the product of three elements along each of the threediagonals going from upper left to lower right; the other three terms carry a negativesign and are products formed along the diagonals from lower left to upper right.Hence,

Db=120312031=120312031123103

=111+220+033-010-321-132.

=1-6-6

Db=-11.

To check this result, we can use Laplace expansion on the elements of the toprow:

Db=120312031=-11+1a11M11+-11+2a12M12+-11+3a13M13

=a11C11-a12C12+a13C13=11231-23201+03103

=1-6-2(3-0)

Db=-11.

For (c), the first row, third row, and the first column all have three zeros out offour terms and so they are all equally good choices for Laplace expansion. Suppose wechoose to expand across the first row, we obtain

Dc=120300302002030030=12-11+1a11M11+-11+2a12M12+-11+3a13M13+-11+4a14M14

=12a11C11-a12C12+a13C13-a14C14

Dc=12-3320003030

Expanding the third-order determinant along the first column, we get

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Dc=-32320003030

=-32-11+1a11M11+-12+1a21M21+-13+1a31M31

=-32a11C11-a21C21+a31C31

=-3230330

=-320-33

=-32-3

=92 .

EXERCISE 3.1.3

By definition, the determinant D2 of the coefficients given pair of equations is

given by

D2=1224=14-22=4-4=0,

while the numerator determinants Nx and Ny are

given by

Nx=3264=34-62=12-12=0,

Ny=1326=16-23=6-6=0.

Because the determinants D2, Nx, and Ny allvanish for the given pair of inhomogeneousequations, solutions may exist but they are notunique. Geometrically, the lines represented by thetwo equations are coincident. That is, if we writeboth of them in slope-intercept form, we obtain thesame equation whose graph is plotted on the xy-coordinate plane below:

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Any ordered pair (x,y) representing any point on the line is a solution of the givensystem. Some of these ordered pairs are tabulated below:

x y

-3 3 y=-12-3+32=32+32

-25

2

y=-12-2+32=22+32

-1 2 y=-12-1+32=12+32

-12

74

y=-12-12+32=14+32

032

y=-120+32=32

1254

y=-1212+32=-14+32

1 1 y=-121+32=-12+32

212

y=-122+32=-22+32

3 0 y=-123+32=-32+32

EXERCISE 3.1.5

(a) Consider two determinants aij and aji, where aij is the general nth orderdeterminant and aji is the nth order determinant formed by interchanging rows andcolumns ofaij. In double-subscript notation,aij=a11a12a1na21a22a2nan1an2ann, andaji=a11a21an1a12a22an2a1na2nann.

Now it is a fundamental theorem that determinants of any order can beevaluated by a Laplace development on any row or column. Thus, we can evaluateboth determinants aij and aji by Laplace expansion either along any row to get

aij=j=1n-1i+jaijMij=j=1naijCij for any i ,

andaji=j=1n-1j+iajiMji=j=1najiCji for any i ,

or along any column to getaij=i=1n-1i+jaijMij=i=1naijCij for any j ,

3

x+2y=3

2y=-x+3

y=-12x+32

2x+4y=6

4y=-2x+6

y=-24x+64

y=-12x+32


4/10

andaji=i=1n-1j+iajiMji=i=1najiCji for any j .

But according to one of the fundamental properties of the nth orderdeterminants, the value of the determinant remains the same if rows andcolumns are interchanged. By this property, we can therefore establish thefollowing:

(i) aij=aji ;

(ii) j=1naijCij=j=1najiCji for any i ;

(iii) i=1naijCij=i=1najiCji for any j .

Now if A=aij , it follows from (i) and (iii) that

iaijCij=iajiCji=A .

(b) Suppose aij and aji are the third-order determinants D3 and D3T :

aij=D3=a11a12a13a21a22a23a31a32a33 and aji=D3T=a11a21a31a12a22a32a13a23a33.

If we form iaijCik where jk, we have the following six cases:

Case 1: j=1, k=2.iaijCik=i=13ai1Ci2=a11C12+a21C22+a31C32

=a11-11+2a21a23a31a33+a21-12+2a11a13a31a33+a31-13+2a11a13a21a23

=-a11a21a33-a23a31+a21a11a33-a13a31-a31a11a23-a13a21

=-a11a21a33+a11a23a31+a21a11a33-a21a13a31-a31a11a23+a31a13a21=0.



=a11a21a32-a22a31-a21a11a32-a12a31+a31a11a22-a12a21

=a11a21a32-a11a22a31-a21a11a32+a21a12a31+a31a11a22-a31a12a21=0.

Case 3: j=2, k=1.

iaijCik=i=13ai2Ci1=a12C11+a22C21+a32C31=a12-11+1a22a23a32a33+a22-12+1a12a13a32a33+a32-13+1a12a13a22a23










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=-a13a21a33-a23a31+a23a11a33-a13a31-a33a11a23-a13a21

=-a13a21a33+a13a23a31+a23a11a33-a23a13a31-a33a11a23+a33a13a21=0.

By exhausting all possible scenarios of jk in third-order determinants (n=3), we have

shown

i=13ai1Ci2=i=13ai1Ci3=i=13ai2Ci1=i=13ai2Ci3=i=13ai3Ci1=i=13ai3Ci2=0.

This process can be carried out, one step at a time, to any n. Therefore, we conclude

iaijCik=0 for jk.

Moreover, analogous to the relationship

iaijCij=iajiCji

established in part (a), we can also write

iaijCik=iajiCki .

It follows that

iaijCik=iajiCki=0 for jk.

EXERCISE 3.1.7

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Method I: CRAMERS RULE

According to Cramers Rule, if the determinant Dn of the coefficients of a set of

linear equations is not zero, then the system has a unique solution, xi=NiDn for1in, where Ni is the determinant obtained by replacing the ith column of Dn by theinhomogeneous terms. Applying Cramers Rule to the given set of six linearinhomogeneous equations with six unknowns yields

x1=N1D=1.00.90.80.40.100.91.00.80.50.20.10.8

0.81.00.70.40.20.70.50.71.00.60.30.60.20.40.61.

00.50.50.10.20.30.51.0D,

x4=N4D=1.00.90.81.00.100.91.00.80.90.20.10.8

0.81.00.80.40.20.40.50.70.70.60.30.10.20.40.61.

00.500.10.20.50.51.0D,

x2=N2D=1.01.00.80.40.100.90.90.80.50.20.10.8

0.81.00.70.40.20.40.70.71.00.60.30.10.60.40.61.

00.500.50.20.30.51.0D,

x5=N5D=1.00.90.80.41.000.91.00.80.50.90.10.8

0.81.00.70.80.20.40.50.71.00.70.30.10.20.40.60.

60.500.10.20.30.51.0D,

x3=N3D=1.00.91.00.40.100.91.00.90.50.20.10.8

0.80.80.70.40.20.40.50.71.00.60.30.10.20.60.61.

00.500.10.50.30.51.0D,

x6=N6D=1.00.90.80.40.11.00.91.00.80.50.20.90.

80.81.00.70.40.80.40.50.71.00.60.70.10.20.40.61

.00.600.10.20.30.50.5D,

whereD=1.00.90.80.40.100.91.00.80.50.20.10.80.81.00.70.40.20.40.50.71.00.60.30.10.20.40.61.00.500.10.20.30.51.0.

Though Cramers Rule is handy for linear systems of two or three equations,using it to solve the given system of six equations with six unknowns is not verypractical.

Method II: GAUSS ELIMINATION

The Gauss elimination technique (also known as echelon method ortriangularization) is a shorter and less complicated way to find the solution of the

given system:

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A0

B0

C0

D0

E0

F0

Since equations A0, B0, C0, D0, E0, and F0 have already been arranged withthe largest coefficients running along the main diagonal, we proceed directly with theelimination process.

Starting off, we use the first equation A0 to eliminate the first unknown x1 fromequations B0, C0, D0, and E0.

A0 : x

1

+0.9

0

x2+0.8

0

x3+0.4

0

x4+ 0.1

0

x5 =1

B1=0.9A0-B0 : - 0.1

9

x2 - 0.0

8

x3 - 0.1

4

x4- 0.1

1

x5- 0.1

0

x6=0

C1=0.8A0-C0 : - 0.0

8

x2 - 0.3

6

x3 - 0.3

8

x4- 0.3

2

x5- 0.2

0

x6=0

D1=0.4A0-D0 : - 0.1

4

x2 - 0.3

8

x3 - 0.8

4

x4- 0.5

6

x5- 0.3

0

x6= -0.3

E1=0.1A0-E0 : - 0.1

1

x2 - 0.3

2

x3 - 0.5

6

x4- 0.9

9

x5- 0.5

0

x6= -0.5

F0 : +0.1

0

x2+0.2

0

x3+0.3

0

x4+ 0.5

0

x5+1.0

0

x6=0.5

Next, we use B1 to eliminate x2 from equations C1, D1, E1, and F0.

B2=-B10.19 :x

2

+0.421052

63

x

3

+ 0.736842

11

x

4

+0.578947

37

x

5

+0.526315

79

x

6

= 0

C2=B2--

C10.08

: - 4.078947

37

x

3

- 4.013157

89

x

4

- 3.421052

63

x

5

- 1.973684

21

x

6

= 0

D2=B2--

D10.14

: -2.29323308

x

3

-

5.26315789

x

4

-

3.42105263

x

5

-

1.61654135

x

6

= -

2.1428571

4

E2=B2--E10.11 : - 2.488038

28

x

3

- 4.354066

99

x

4

- 8.421052

63

x

5

- 4.019138

76

x

6

= -

4.5454545

5

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F1=B2-F00.10 : - 1.578947

37

x

3

- 2.263157

89

x

4

- 4.421052

63

x

5

- 9.473684

21

x

6

= -5

Repeating the same technique above, we use C2 to eliminate x3 from equations D2,E2, and F1.C3=-C24.078 :x

3

+ 0.9838709

68

x

4

+0.8387096

77

x

5

+ 0.4838709

68

x

6

=0

D3=C3--

D22.293

: - 1.3112109

99

x

4

- 0.6530936

01

x

5

- 0.2210470

65

x

6

= -0.93442623

E3=C3--

E22.488

: - 0.7661290

32

x

4

- 2.5459057

07

x

5

- 1.1315136

48

x

6

= -

1.82692307

7

F2=C3--

F11.578

: - 0.4494623

66

x

4

- 1.9612903

23

x

5

- 5.5161290

32

x

6

= -

3.16666666

7

We eliminate the remaining unknowns x4, x5, and x6 similarly.

D4=-D31.311 :x

4

+0.4980842

91

x

5

+ 0.1685823

75

x

6

=0.71264367

8

E4=D4--

E30.766

: - 2.8249926

32

x

5

- 1.3083407

01

x

6

= -

1.671971706

F3=D4--

F20.449

: - 3.8655520

72

x

5

- 12.104144

90

x

6

= -

6.33281086

7

E5=-E42.824 :x

5

+0.4631306

60

x

6

=0.59184993

5

F4=E5--

F33.865

: - 2.6681542

56

x

6

=-

1.04641822

1

F5=-F42.668 :x

6

=0.3921880

52

We obtain the values of the other unknowns by working back up. Starting from E5 , we

get

x5+0.463130660x6=0.591849935

x5=0.591849935-0.463130660x6

x5=0.591849935-0.4631306600.392188052=0.410215624.

Substituting these values of x5 and x6 to D4 yields

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x4+0.498084291x5+0.168582375x6=0.712643678

x4=0.712643678-0.498084291x5-0.168582375x6

x4=0.712643678-0.4980842910.410215624-0.1685823750.392188052=0.442205726.

Then from C3 , we have

x3+0.983870968x4+0.838709677x5+0.483870968x5=0

x3=-0.983870968x4-0.838709677x5-0.483870968x5

x3=-0.9838709680.442205726-0.8387096770.410215624-0.4838709680.392188052

x3=-0.968893602;

from B2, we have

x2+0.42105263x3+0.73684211x4+0.57894737x5+0.52631579x6=0

x2=-0.42105263x3-0.73684211x4-0.57894737x5-0.52631579x6

x2=-0.42105263-0.968893602-0.736842110.442205726-0.578947370.410215624-0.526315790.392188052x2=-0.361788618;

and finally from A0, we have

x1+0.90x2+0.80x3+0.40x4+0.10x5=0

x1=-0.90-0.361788618-0.80-0.968893602-0.400.442205726-0.100.410215624x1=1.882820785.

Therefore, the solution of the given set of six linear inhomogeneous equations with six

unknowns is:

x1=1.88282;

x2=-0.36179;

x3=-0.96889;

x4=0.44221;x5=0.41026;

x6=0.39219.

EXERCISE 3.1.9

By definition of the dot product, the given vector equations can be transformed

into the following scalar equations:

ax=d

:a1x1+a2x2+a3x3=d

bx=e

:b1x1+b2x2+b3x3=e

cx=f

:c1x1+c2x2+c3x3=f

wherex=x1i+x2j+x3k ,

a=a1i+a2j+a3k ,

b=b1i+b2j+b3k ,

c=c1i+c2j+c3k .

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Now according to Cramers Rule, this resulting system of linear equations has a

unique solution

x1=N1Dn=da2a3eb2b

3fc2c3Dn

x2=N2Dn=a1da3b1eb

3c1fc3Dn

x3=N3Dn=a1a2db1b2

ec1c2fDn

if the determinant Dn of the coefficients is not zero: Dn=a1a2a3b1b2b3c1c2c30 .

By definition of the triple scalar product,

Dn=a1a2a3b1b2b3c1c2c3=abc ,

with abc0, as specified in the problem.

Since Dn0, the system indeed has a unique solution given by

x1=N1abc x2=N2abc x3=N3abc

where the numerator determinants N1, N2, N3 can be evaluated using Laplaceexpansion along the columns containing the constants d, e, f. Hence, for N1 we

expand along the first column to obtain

N1=da2a3eb2b3fc2c3=d(-1)1+1b2b3c2c3+e(-1)2+1a2a3c2c3+f(-1)3+1a2a3b2b3

=db2c3-b3c2-ea2c3-a3c2+fa2b3-a3b2

N1=dbci-eaci+fabi .

Then for N2 and N3 , we expand along the second and third column, respectively:

N2=a1da3b1eb3c1fc3=d(-1)1+2b1b3c1c3+e(-1)2+2a1a3c1c3+f(-1)3+2a1a3b1b3

=-db1c3-b3c1+ea1c3-a3c1-fa1b3-a3b1

N2=dbcj-eacj+fabj .

N3=a1a2db1b2ec1c2f=d(-1)1+3b1b2c1c2+e(-1)2+3a1a2c1c2+f(-1)3+3a1a2b1b2

=db1c2-b2c1-ea1c2-a2c1+fa1b2-a2b1

N3=dbck-eack+fabk .

Using the results obtained from Laplace development by complementary minors,the solution of the system of linear equation becomes

x1=dbci-eaci+fabiabc ,

x2=dbcj-eacj+fabjabc ,

x3=dbck-eack+fabkabc .

The ordered triple x1, x2, x3 can also be expressed in vector form as

x=x1, x2, x3

=x1i+x2j+x3k

=dbci-eaci+fabiabci+dbcj-eacj+fabjabcj+dbck-eack+fabkabck

=dbcii+bcjj+bckkabc-eacii+acjj+ackkabc+fabii+abjj+abkkabc

x=dbc-eac+fababc=dbc+eca+fababc .

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answers to problem set # 7

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