LINEAR ALGEBRA
Many times we have toencounter the situations ofsolving system of linearequations in several variables.
For example.2x + y+ 2z - 12u + v + 2w = 1
x + 2y – z + 6u + 2v – w = 2
5x + 4y + 3z + 45u + 4v+ 3w= 4
3x + 20y - 2z - 6u + 8v - 4 w=2
20x + y - 2z - 2u + v + 2w = 7
2x - 3y + 20z + 5u + 4v - 5w = 19
QUESTIONS :
1)Whether the solution exists?
2)If the solution exists then whether there
exist(s) ,
(a) Only one solution (UNIQUE)?
(b) More than one solutions?
(c) Infinite solutions?
and
(d)Can we have a simple method ?
Cramers Rule could not answer
(a) (b) (c) (d).
Rank of a matrix helps the methods, which
Answer (a) (b) (c) (d) in a far better way
than Cramers Rule.
Definition: Rank of a matrix A is rif,
i) it has at least one non-zerominor of order r andii) all the minors of order higher
than r are zeroes.
Notation. If Rank of a matrix Ais r then it is denoted as (A) = r
Elementary Transformations
The following operations w.r.t. a matrix are known
as elementary transformations.
Inter change of any two ROWS, indicated by Rij.
Multiplication of elements of any ROW by a
non-zero real number, indicated by kRi
Addition of the constant multiple of ith ROW to
jth ROW indicated by Ri + kRj
Similar COLUMN transformation are denoted by
Cij , kCi, Ci + kCj
Example. Using elementary transformations
reduce the matrix A first to upper triangular and
then to identity matrix where
A =
213
321
752
Solution:
213
321
752
Operating R1 R12
A =
213
752
321
Operating R2 R2-2R1, R3 R3-3R1
A~
750
110
321
Operating R2 R3 R3 +5R2
A~
200
110
321
(Upper Triangular
Form)
Operating C2 C2-2C1, C3 C3-3C1
A~
200
110
001
Operating C3 C3-C2
A~
200
010
001
Operating C3 C3/(-2)
A~
100
010
001
(Identity matrix)
Definition:Two matrices A and B are said to
be eqivalent if one can be obtained from another
by a sequence of elementary transformations and
the same is symbolically written as A~B.
Example. Using elementary transformations
find the ranks of the following matrices
(1)
7036
2313
4211
1132
(2)
1063
742
321
(3)
8
0
0
201
342
121
(4)
12
9
3
1
6
7
1
3
3
6
3
4
1
3
2
1
SOLUTIONS
(1)Let A =
7036
2313
4211
1132
Operating R12
7036
2313
1132
4211
Operating R2R2 -2R1, R3R3-3R1, R4 R4-6R1
Operating R2 R2 -R3,
171290
10940
7350
4211
171290
10940
3610
4211
Operating R3 R3-4R2 , R4 R4-9R2
446600
223300
3610
4211
Operating , R4 R4-2R3
A ~
0000
223300
3610
4211
Operating R2R2 -2R1, R3R3-3R1, R4 R4-6R1
Observe that the matrix is in upper
TRIANGULAR FORM
The rank of the matrix = the number of non zero
rows (A) = 3
B =
1063
742
321
Operating R2R2 -2R1, R3R3-3R1
B ~
100
100
321
though the matrix is in upper TRIANGULAR
FORM we observe that
R2 and R3 are identical.
R3R3-R2
B ~
000
100
321
This form of the
matrix where,
(i)First entry in every non-zero row is ‘UNITY’.
(ii)Elements in the column below these ‘above
mentioned’ UNITIES are zeroes
(iii)All the zero rows are below the non-zero rows is
known as ROW ECHELON FORM
The rank of the matrix = the number of non zero
rows (A) = 2
(3) Reduce to row -echelon form and find rank
of C for,
C=
8
0
0
201
342
121
Operating R2 R2+2R1, R3 R3-R1
C~
8
0
0
120
580
121
Operating R23
C~
0
8
0
580
120
121
Operating R3 R3+4R2
Operating C ~
0
8
0
900
120
121
Operating)2(
2
C
C ~
0
8
0
900
110
111
Operating9
3R
C ~
0
8
0
100
110
111
This is in
Row- echelon form.
The rank of the matrix = the number of non zero
rows (A) = 3
(4)
C=
12
9
3
1
6
2
4
2
6
7
1
3
3
6
3
4
3
1
2
1
Operating C2 C2-4C1, C3 C3-3C1
C4 C4+2C1 C5 C5-C1
C~
15
10
5
0
15
10
5
0
15
10
5
0
15
10
5
0
1
3
2
1
Operating C3 C3 -C2 C4 C4 -C2
C5 C5 -C2
C~
0
0
0
0
0
0
0
0
0
0
0
0
15
10
5
0
1
3
2
1
Upper Triangular form
The rank of the matrix = the number of non zero
coloumns (A) = 2
(5) Find the rank by reducing it to the
following matrix to normal form
D~
98
97
96
95
97
96
95
94
96
95
94
93
95
94
93
92
94
93
92
91
Operating R2R2 -R1, R3R3 -R1 , R4R4 -R1
D~
3
2
1
95
3
2
1
94
3
2
1
93
3
2
1
92
3
2
1
91
Operating R12
D~
3
2
95
1
3
2
94
1
3
2
93
1
3
2
92
1
3
2
91
1
Operating R2R2 -91R1, R3R3 -2R1 ,
R4R4 -3R1
D~
0
0
4
1
0
0
3
1
0
0
2
1
0
0
1
1
0
0
0
1
C1C1 –C2 , C3 C3 -C2 , C4 C4 - C2
D~
0
0
3
0
0
0
2
0
0
0
1
0
0
0
1
1
0
0
0
0
D~
3
2
1
95
3
2
1
94
3
2
1
93
3
2
1
92
3
2
1
91
CONSISTENCY OF SYSTEM OFSIMULTANEOUS LINEAR EQUATIONS
Example 1.Test the consistency and solve
x-2y+3z=82x-3y=-5 ( *)
x+y+z=9
Solution: Consider The Augumented matrix of the system (*)
We findi)the rank of matrix Aii) the rank of matrix [A/D]
If
A =
111
032
321
X =
z
y
x
D =
9
5
8
Then (*) in matrix form is
AX = D
Consider [A/D]=
9
5
8
1
0
3
1
3
2
1
2
1
(Augumented matrix)
Operating R2R2 -2R1, R3R3-R1
[A/D] ~
1
21
8
2
6
3
3
1
2
0
0
1
Operating R3R3-3R2
[A/D] ~
64
21
8
16
6
3
0
1
2
0
0
1
(**)
( (A) = 3 ) = ((A/D) = 3)
System of equations (*) are consistent.
( (A) = 3 ) = ((A/D) = 3)= The of unknowns
There exists a unique solution of (*)
(*) and (**) are equivalent system
x - 2y + 3z = 8 --------(i)
(**) y - 6z = -21--------(ii)
16z = 64--------(iii)
(iii) z = 4
z = 4 in (ii) y = 3
y = 3 and z = 4 in (i) x = 2
Answer x = 2, y = 3 and z = 4
Example 2.
Test the consistency and solve
2x+y+2z=1x+2y - z=2 ( *)
5x+4y+3z=4
Solution: Consider The Augumented matrix of the system (*)
We findi)the rank of matrix Aii) the rank of matrix [A/D]
If
A =
345
121
212
X =
z
y
x
D =
4
2
1
Then (*) in matrix form is
AX = D
Consider[A/D]=
4
2
1
3
1
2
4
2
1
5
1
2
(Augumented matrix)
Operating R12
[A/D] ~
4
1
2
3
2
1
4
1
2
5
2
1
Operating R2R2 -2R1, R3R3-5R1
[A/D] ~
6
3
2
8
4
1
6
3
2
0
0
1
Operating R3R3-2R2
[A/D] ~
0
3
2
0
4
1
0
3
2
0
0
1
(**)
( (A) = 2 ) = ((A/D) = 2)
System of equations (*) are consistent.
( (A) = 2 ) = ((A/D = 2) < The of unknowns=3
There exist infinite solutions of (*)
(*) and (**) are equivalent system
x+ 2y - z = 2 --------(i)
(**) -3y+ 4z = -3--------(ii)
Let z = k be the parameters
z = k in (ii) y = 1+34
k
y = 1+34
k and z = k in (i) x = -35
k
Answer: x = -35
k y = 1+34
k z = k
For example k=3 x = -5 y = 5 z = 3
is one particular solution.
Example 3.
Test the consistency and solve
x-4y+7z=143x+8y -2 z=13 ( *)
7x-8y+26z=5
Solution: Consider The Augumented matrix of the system (*)
We findi)the rank of matrix Aii) the rank of matrix [A/D]
If
Consider[A/D]=
5
2
14
26
2
7
8
8
4
7
3
1
Operating R2R2 -3R1, R3R3-7R1
[A/D] ~
93
29
14
23
23
7
20
20
4
0
0
1
Operating R3R3-R2
[A/D] ~
64
29
14
0
23
7
0
20
4
0
0
1
( (A) = 2 ) ((A/D) = 3)
There exists no solution of system ofsimultaneous linear equations(*).
GAUSS-SEIDAL ITERATION ETHOD
(of solving the system linear
simultaneous equations.)
Example 1.Use Gauss-Seidal iteration method to
solve
the following system of equations.
3x + 20y - 2z = -18
20x + y - 2z = 17 (A)
2x - 3y + 20z = 25
Solution. Rearranging the system of equations (A)
20x + y - 2z = 1714131211
azayaxa
3x + 20y - 2z = -1824232221
azayaxa (B)
2x - 3y + 20z =2534333231
azayaxa
The above system equations is arranged such
that,
20 > 1 + 111
a >12
a +13
a
20 > 3 + 222
a >21
a +23
a
20 > 1 + 133
a >31
a +32
a
PROCESS of rearranging the system equations
satisfying the above conditions, is known as
DIAGONALIZATION of equations and the
SYSTEM is known as DIAGONALY
DOMINANT.
System (B)
x =201
[ 17- y +2z ]
y =201
[-18-3x + z ] (C)
z =201
[25 –2x + 3y ]
Let the initial approximations to the solution of the
system (A) be
,0)0( x ,0)0( y 0)0( z
First Iteration:Using ( C )
)1(x201
[ 17- y )0( +2z )0( ]
)1(x201
[ 17- 0 + 2(0) ] = 8500.0
y )1( =201
[-18-3x )1( + z )0( ]
y )1( =201
[ -18-3(0.8500 ) + 0 ] = -1.0275
z )1( =201
[25 –2x )1( + 3y )1( ]
z )1( =201
[25 –2(0.8500) + 3(- 1.0275)]
= 1.0109
Second Iteration: Usining ( C )
)2(x201
[ 17- y )1( +2z )1( ]
)2(x201
[ 17- (-10275) + 2(1.0109 ) ]
= 1.0025
y )2( =201
[-18-3x )2( + z )1( ]
y )2( =201
[ -18-3(1.0025 ) + 1.0109 ]
= - 0.99928
z )2( =201
[25 –2x )2( + 3y )2( ]
z )2( =201
[25 –2(1.0025) + 3(- 0.99928)]
= 0.9998
Third Iteration: Usining ( C )
)3(x201
[ 17- y )2( +2z )2( ]
)3(x201
[ 17- (- 0.99928)+ 2(0.9998) ]
= 1.0000
y )3( =201
[-18-3x )3( + z )2( ]
y )3( =201
[ -18-3(1.0000) +0.9998 ]
= -1.0000
z )3( =201
[25 –2x )3( + 3y )3( ]
z )3( =201
[25 –2(1.00) + 3(-1.0000)]
= 1.0000
Answer after three iterations
x =1.000, y = -1.0000 and z = 1.000
Example 2.Use Gauss-Seidal iteration method to
solve
the following system of equations.
x + y +54z=110
27x +6 y - z = 85 (A)
6x+15y + 2z =72
Solution. Rearranging the system of equations (A)
27x +6 y - z = 8514131211
azayaxa
6x+15y + 2z =7224232221
azayaxa (B)
x + y +54z=11034333231
azayaxa
The above system equations is arranged such
that,
27 > 6 + 111
a >12
a +13
a
15 > 6 + 222
a >21
a +23
a
54 > 1 + 133
a >31
a +32
a
System (B)
x =271
[ 85-6 y +z ]
y =151
[72-6x -2z ] (C)
z =541
[110 –x -y ]
Let the initial approximations to the solution of the
system (A) be
,1)0( x ,0)0( y 0)0( z
First Iteration:Using ( C )
)1(x271
[ 85-6 y )0( +z )0( ]
)1(x271
[ 85- 6 (0) +0] =3.148148
y )1( =151
[72- 6x )1( -2 z )0( ]
y )1( =151
[72- 6(3.148148) -2 (0) ] =3.54074
z )1( =541
[110 –x )1( -y )1( ]
z )1( =541
[110 –3.148148 -3.54074 ]=1.913168
Second Iteration: Usining ( C )
)2(x271
[ 85-6 y )1( +z )1( ]
)2(x271
[ 85- 6 (3.54074) +1.913168]
=2.432175
y )2( =151
[72- 6x )2( -2 z )1( ]
y )2( =151
[72- 6(2.432175) -2( 1.913168) ]
= 3.57204
z )2( =541
[110 –x )2( -y )2( ]
z )2( =541
[110 –2.432175 -3.57204 ]
=1.925837
ert45t
Third Iteration: Usining ( C )
)3(x271
[ 85-6 y )2( +z )2( ]
)3(x271
[ 85- 6 (3.57204) +2(1.925837)]
=2.425689
y )3( =151
[72- 6x )3( -2 z )2( ]
y )3( =151
[72- 6(2.425689) -2( 1.913168) ]
=3.57313
z )3( =541
[110 –x )2( -y )2( ]
z )3( =541
[110 –2.432175 -3.57204 ]
=1.925947
Answer after three iterations
x =2.425689, y = 3.57313 and z = 1.925947
EIGEN VALUESAND
EIGENVECTORSOF A MATRIX
Definition: Let A be a nn square matrix.then,i)Determinant of (A- I ) = 0 or Symbolically
0 IA is known as characteristic equation
of the matrix Aii) Roots of the characteristic equation are
known as characteristic rots or eigen valuesof the matrix A
iii) X, The non-trivial solution of the system(A- I )X = 0 is known as characteristic vector
or eigenvector of the matrix A w.r.t.eigenvalue
EXAMPLES:Example (1). Find the eigenvalues and eigenvectorsof the given matrices.
i) A=
342
476
268
Characteristic equation is IA 0
342
476
268
= 0
34
47)8(
32
466
42
762
=0
Expandining the determinant we get
3 218 045 OR 2( 18 0)45 as the characteristic equation
Solvining this characteristic equation we get 3 roots
of the equation 0,3,15 and are known as
characteristic roots or eigen values of the matrix A
EIGENVECTORS: Let X=
z
y
x
be the eigenvector
of matrix A corresponding to eigenvalue
(A- I) X = 0
342
476
268
z
y
x
=
0
0
0
For = 0
342
476
268
z
y
x
=
0
0
0
Operating R13
268
476
342
z
y
x
=
0
0
0
Operating R2 R2 +3 R1, R3 R3 - 4 R1
10100
550
342
z
y
x
=
0
0
0
Operating R3 R3 + 2 R2
000
550
342
z
y
x
=
0
0
0
The above system is equivalent to2x-4y+3z = 0
-5y+5z = 0
which generate the solutions as
z
y
x
=
k
k
k
2
2
eigenvector of matrix A corresponding to eigenvalue =0for example if k=1
z
y
x
=
2
2
1
For = 3 we get
042
446
265
z
y
x
=
0
0
0
Operating R1R1 + R2
042
446
221
z
y
x
=
0
0
0
Operating R2 R2 -6 R1, R3 R3 + 2 R1
480
8160
221
z
y
x
=
0
0
0
Operating R3 R3 +2
1R2
000
8160
221
z
y
x
=
0
0
0
The above system is equivalent tox+2y+2z = 0
16y+8z = 0
which generate the solutions as
z
y
x
=
k
k
k
2
2
eigenvector
of matrix A corresponding to eigenvalue =3for example if k=-1
z
y
x
=
2
1
2
For = 15 we get
1242
486
267
z
y
x
=
0
0
0
Operating R1R1 - R2
1242
486
621
z
y
x
=
0
0
0
Operating R2 R2 -6 R1, R3 R3 + 2 R1
000
40200
621
z
y
x
=
0
0
0
The above system is equivalent to-x+2y+6z = 0
-20y-40z = 0
which generate the solutions as
z
y
x
=
1
2
2
k
k
eigenvector
of matrix A corresponding to eigenvalue =15for example if k =1
z
y
x
=
1
2
2
Example (2)
A=
113
151
311
Characteristic equation is IA 0
113
151
311
= 0
11
15)1(
13
11)1(
13
513
=0
Expanding the determinant we get
3 27 036 as the characteristic equation
-2, 3, 6 are the eigen values of the matrix A.
EIGENVECTORS: Let X=
z
y
x
be the eigenvector
of matrix A corresponding to eigenvalue
(A- I) X = 0
113
151
311
z
y
x
=
0
0
0
For = -2
313
171
313
z
y
x
=
0
0
0
R3 R3 - R1
000
171
313
z
y
x
=
0
0
0
Operating R12
000
313
171
z
y
x
=
0
0
0
Operating R2 R2 -3 R1
000
0200
171
z
y
x
=
0
0
0
The above system is equivalent toX+7y+z = 0
-20y = 0
which generate the solutions as
z
y
x
=
k
k
0 = k
1
0
1
eigenvector of matrix A corresponding to eigenvalue =0for example if k=3.43
z
y
x
=
43.3
0
43.3
For = 3 we get
213
121
312
z
y
x
=
0
0
0
Operating R12
213
312
121
z
y
x
=
0
0
0
Operating R2 R2 +2 R1, R3 R3 -3R1
550
550
121
z
y
x
=
0
0
0
Operating R3 R3 + R2
000
550
121
z
y
x
=
0
0
0
which generate the solutions as
z
y
x
=
k
k
k
eigenvector
of matrix A corresponding to eigenvalue =3
for example if k = 1
z
y
x
=
1
1
1
For = 6 we get
513
111
315
z
y
x
=
0
0
0
Operating R12
513
315
111
z
y
x
=
0
0
0
Operating R2 R2 +5R1, R3 R3 -3 R1
840
840
111
z
y
x
=
0
0
0
Operating R3 R3 + R1
000
840
111
z
y
x
=
0
0
0
The above system is equivalent to
x-y+z = 0-4y+8z = 0
which generate the solutions as
z
y
x
=
k
k
k
2 eigenvector
of matrix A corresponding to eigenvalue =6for example if k=1
z
y
x
=
1
2
1
Example (3)
A=
021
612
322
Characteristic equation is IA 0
21
612
322
= 0
1
61)2(
1
622
21
123
=0
Expanding the determinant we get
3 212 -45=0as the characteristic equation
-3,-3,5 are the eigen values of the matrix A.
EIGENVECTORS: Let X=
z
y
x
be the eigenvector
of matrix A corresponding to eigenvalue
(A- I) X = 0
21
612
322
z
y
x
=
0
0
0
For = -3
321
642
321
z
y
x
=
0
0
0
Operating R12
321
642
321
z
y
x
=
0
0
0
Operating R2 R2 -2 R1, R3 R3 + R1
000
000
321
z
y
x
=
0
0
0
The above system is equivalent to
x+2y-3z = 0
Let z=k1y=k
2
which generate the solutions as
z
y
x
=
1
2
2123
k
k
kk
For = 5
521
642
327
z
y
x
=
0
0
0
Operating R13
327
642
521
z
y
x
=
0
0
0
Operating R2 R2 +2 R1, R3 R3 -7 R1
32160
1680
521
z
y
x
=
0
0
0
Operating , R3 R3 +2 R2
000
1680
521
z
y
x
=
0
0
0
which generate the solutions as
z
y
x
=
k
k
k
2
9
