answers to problem set # 7
TRANSCRIPT
-
8/14/2019 Answers to Problem Set # 7
1/10
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
1
-
8/14/2019 Answers to Problem Set # 7
2/10
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:
2
-
8/14/2019 Answers to Problem Set # 7
3/10
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
-
8/14/2019 Answers to Problem Set # 7
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.
Case 2: j=1, k=3.iaijCik=i=13ai1Ci3=a11C13+a21C23+a31C33
=a11-11+3a21a22a31a32+a21-12+3a11a12a31a32+a31-13+3a11a12a21a22
=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
=a12a22a33-a23a32-a22a12a33-a13a32+a32a12a23-a13a22
=a12a22a33-a12a23a32-a22a12a33+a22a13a32+a32a12a23-a32a13a22=0.
Case 4: j=2, k=3.iaijCik=i=13ai2Ci3=a12C13+a22C23+a32C33
=a12-11+3a21a22a31a32+a22-12+3a11a12a31a32+a32-13+3a11a12a21a22
=a12a21a32-a22a31-a22a11a32-a12a31+a32a11a22-a12a21
=a12a21a32-a12a22a31-a22a11a32+a22a12a31+a32a11a22-a32a12a21=0.
Case 5: j=3, k=1.iaijCik=i=13ai3Ci1=a13C11+a23C21+a33C31
=a13-11+1a22a23a32a33+a23-12+1a12a13a32a33+a33-13+1a12a13a22a23
=a13a22a33-a23a32-a23a12a33-a13a32+a33a12a23-a13a22
4
-
8/14/2019 Answers to Problem Set # 7
5/10
=a13a22a33-a13a23a32-a23a12a33+a23a13a32+a33a12a23-a33a13a22=0.
Case 6: j=3, k=2.iaijCik=i=13ai3Ci2=a13C12+a23C22+a33C32
=a13-11+2a21a23a31a33+a23-12+2a11a13a31a33+a33-13+2a11a13a21a23
=-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
5
-
8/14/2019 Answers to Problem Set # 7
6/10
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:
6
-
8/14/2019 Answers to Problem Set # 7
7/10
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
7
-
8/14/2019 Answers to Problem Set # 7
8/10
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
8
-
8/14/2019 Answers to Problem Set # 7
9/10
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 .
9
-
8/14/2019 Answers to Problem Set # 7
10/10
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 .
10