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Contents abstract:............................................................................................................................................1 keywords:.........................................................................................................................................2 introduction:.....................................................................................................................................2 Review of Literature:.......................................................................................................................2 The linear homogeneous system of matrices:..................................................................................2 The linear non-homogeneous system of matrices...........................................................................3 Gauss elimination method:..............................................................................................................5 Gauss-Jordon elimination method:..................................................................................................5 L.U. decomposition method:...........................................................................................................6 Doolittle’s method:..........................................................................................................................7 General formulation of Doolittle’s method......................................................................................7 Crout’s method.................................................................................................................................8 CONCLUSION:..............................................................................................................................9 References........................................................................................................................................9

ABSTRACT: Matrices are used to solve any linear equations. Linear equations are divided into homogenous and non-homogeneous systems. ( Ax = B ) is the rule that we apply on it using the matrix where A is the coefficient of the variables and x is the variable and ( B∈R ) is the result of the equations. The difference between homogenous and non-homogeneous systems is in B as is in homogenous system the results equals to zero while non-homogeneous the result is (∈R{0}). And will be solved by two theorems which are Crout’s method and Doolittle’s method, and we mentioned the difference between the two methods.

KEYWORDS: Homogeneous, Trivial, Non-trivial, Non-homogeneous, Gauss, Gauss Jordan, Lower-Upper, Doolittle and Crout.

INTRODUCTION: The matrix is the easiest way to show how to solve the ordinary differential equation. as it plays a very important rule in physics and engineering fields; before founding the matrix, the equations were written in a way that the equations are separated but after the matrix it is written in one matrix and solved easily.

1

REVIEW OF LITERATURE: The linear homogeneous system of matrices: The system has all the numbers that are on the right of equations are zeros, and have a single solution and is called the easy solution, that if we put the value of all variables worthy zero when they collected them the right party of all equations is equal to zero ... and this is achieved when it is (number of unknowns = number of equations) So what happens when the number of unknowns is the greater the number of equations? Theorem: Free Variable for Homogeneous System: If there is a homogeneous linear system has a number of variables (N) and if their extended matrix (R) of non-zero rows, so this system (N-R) of free variations. Easy example: If we have 3 equations, 4 unknowns, what we can do to produce the number of free variations? Solution: (numbers of unknowns - numbers of equations = numbers of free variations) Theorem: (when the number of unknowns greater than the number of equations) Example: So the No. of unknown ˃ the no. of equation. 3P1 + P2 + P3 + P4 = 0 , 5P1 - P2 + P3 – P4 = 0 3 1 1 1 ⋮ 0 8 0 2 0 ⋮ 0 1/8 R1 R2+R1 R1 5 −1 1 −1 ⋮ 0 5 −1 1 −1 ⋮ 0 1 0 1 /4 0 ⋮ 0 1 0 1/4 0 ⋮ 0 -5R1+R2 R2 -R2 5 −1 1 1 ⋮ 0 0 −1 −1/ 4 −1 ⋮ 0 1 0 1/4 0 ⋮ 0 0 1 1/4 1 ⋮ 0 {rref} We have 2 free variables 1 P3 = s P2 + P3 + P4 = 0 4 1 s + t= 0 , P2 = P4 = t , (s,t) ∈ R P2 + 4 1 P1 + P3 = 0 , P1 = 4

[ [ [

]

] ]

[

]

[

]

1 s-t 4 1 s 4

Note: REF: we have some properties to put the matrix in the row echelon form: 1. All zero rows should appear below the matrix. 2. Every non-zero row must have the first element in it is 1 to the left of this row and this element is called leading 1. 3.

Any two non-zero consecutive rows the leading 1 of the top row should appear to the left of the leading 1 of the bottom row.  When the zeros are found on left and right of the leading 1, so the matrix become to reduced echelon form. Another explanation: We have two options, but we have an equation to told us which option that we must use: Unique or trivial sol. Y=0

(AY = 0)

2

infinite or nontrivial sol. Y≠0

Example for trivial solution: H1 – 2H2 + H3 = 0

,

6H2 – 3H3= 0

[

1 −2 1 ⋮ 0 0 6 −3 ⋮ 0 1 −2 −1 ⋮ 0

Sol. :

]

-R1 +R3

,

H1 - 2H2 - H3 = 0 R3 ,

1 R2 3

[

1 −2 1 ⋮ 0 0 2 −1 ⋮ 0 0 0 −2 ⋮ 0

]

H3 = 0 H2 = 0 2H2 –H3 = 0 H1 – 2H2 + H3 = 0 H1 = 0 (0,0,0) Trivial solution Example for non-trivial solution: L1 + 2L2 + 3L3 + 2L4 = 0 (a) , L1 + 3L2 + 5L3 + 5L4 = 0 (b) , L1 + 2L2 + 3L3 + 2L4 = 0 (c) , -L1 – 2L2 – 6L3 + 7L4 = 0 (d)

[

1 2 3 1 3 5 1 2 3 −1 −2 −6

-2R2+R1

[

1 0 0 0

0 1 0 0

2 5 2 7

] [

⋮ ⋮ ⋮ ⋮

0 0 0 0

1 0 0 0

−1 R1 , R4 3

0 −7 ⋮ 0 0 9 ⋮ 0 1 −3 ⋮ 0 0 0 ⋮ 0

Sol.

]

–R1+R2

R2 , -2R1+R3

0 −1 −4 ⋮ 1 2 3 ⋮ 0 1 −3 ⋮ 0 1 −3 ⋮

R3 , R1+R4

0 0 0 0

]

R3+R1

R4

[

2 3 2 ⋮ 1 2 3 ⋮ 0 1 −3 ⋮ 0 −3 9 ⋮

1 0 0 0

R1 , -2R3+R2

R2 , -R3+R4

0 0 0 0

]

R4

(So the No. of unknown ˃ the no. of equation)

L1 – 7L4 = 0

,

L2 + 9L4 = 0

,

L3 – 3L4 = 0

(L4) is a free variable, L4=k.L1 =7k , L2 = -9k , L3 =3k Solution: (7k , -9k , 3k , k) A particular solution: (k = 1), (7 , -9 , 3 , 1), and use these to solve a,b,c,d equations to prove that equal ZERO.

The linear non-homogeneous system of matrices It is a system made up of a straight-line equation, and the numbers to the right of the equation are numbers that are not equal to zero in all equations. So it is possible to have two equations or an equation equal to zero when solving with matrices.

3

There are 3 types of non-homogeneous system: 1. Unique solution. 2. Infinite solution. 3. No solution. Non-homogenous (Heterogeneous) system b11X + b12Y + b13Z =a1 b11 b12 b 13 X a1 = b b b a2 Y b21X + b22Y + b23Z =a2 21 22 23 b31 b32 b 33 Z a3

[

][ ] [ ]

b31X + b32Y + b33Z =a3 −1 −1 −1 ⸪AX = B B ⸪ X = A−1 B ¿A A AX = A Example: Q1 + 2Q2 + Q3 = 4 , 3Q1 – 4Q2 – 2Q3 = 2 , 5Q1 + 3Q2 + 5Q3 = -1 Solve the equation using matrix. Solution: X = A−1 B 1 2 1 X 4 1 adj ( A ) A−1= = 3 −4 −2 Y 2 | A| −1 5 3 5 Z

[

][ ] [ ] | | [ ] [ ] [] [ ][ ] [ ] [ ] 1

2

1

5

3

5

|A|= 3 −4 −2 =−14−50+29 =−35

⸪Coff. (A) =

−1 ⸪ A =

−14 −25 29 −7 0 7 0 5 −10

0 −1 −14 −7 −25 0 5 35 29 7 −10

⸪

adj ( a )=[ Coff . ( A )]

⸪X =

t

=

[

−14 −7 0 −25 0 5 29 7 −10

]

−1

A B

0 X −1 −14 −7 4 −1 −70 2 2 = ⸪ Y = −25 0 5 −105 = 3 35 35 Z 140 29 7 −10 −1 −4

X =2, Y =3 , Z =−4

Definition of the RANK: It is the number of non-zero rows after first row operations. As we mentioned, the non-homogeneous system has three states: 1) A unique solution IF R(A)=R(A\B) =No. of unknown. Example Q1 - 2Q2 + 3Q3 = 9 , 2Q1 – 5Q2 + 5Q3 = 17 , -1Q1 + 3Q2 = -4 1 −2 3 ⋮ 9 1 −2 3 ⋮ 9 Solution: -2R1+R2 R2 , R1+R3 R3 2 −5 5 ⋮ 17 0 −1 −1 ⋮ −1 0 1 3 ⋮ 5 −1 3 0 ⋮ −4

[

R2

[

R3

]

[

1 −2 3 ⋮ 9 0 1 3 ⋮ 5 0 −1 −1 ⋮ −1

]

R2 + R3

]

R3

[

1 −2 3 ⋮ 9 0 1 3 ⋮ 5 0 0 1 ⋮ 2 By back substation Z = 2 , Y = -1 , X = 1 ⸪R(A)=R(A\B) = no. of unknown = 3 ⸪ unique solution

4

[

1 −2 3 ⋮ 9 0 1 3 ⋮ 5 0 0 2 ⋮ 4

]

]

R3 2

2) An infinite solution IF no. of unknowns > the no. of the equations. Example: Q1 + 2Q2 - 3Q3 = 4 , 3Q1 – 1Q2 + 5Q3 = 2 , 4Q1 + 1Q2 + 2Q3 = 6 Solution: 1 2 −3 ⋮ 4 1 2 −3 ⋮ 4 R2 , -4R1+R3 R3 3 −1 5 ⋮ 2 -3R1+R2 0 −7 14 ⋮ −10 4 1 2 ⋮ 6 0 −7 14 ⋮ −10 R3 1 2 −3 ⋮ 4 1 2 −3 ⋮ 4 −R 2 10 R2 0 1 −2 ⋮ 0 −7 14 ⋮ −10 7 7 0 0 0 ⋮ 0 0 0 0 ⋮ 0

[ [

]

[

[

]

]

⸪R(A)=R(A\B) < number of unknown ⸪The number of solution is: infinite 3) No solution IF R(A) No. of unknown

a) no solution

b) unique solution

c) infinite solution

10) Which of the following used by back substitution?

10

d) none of the above

a) Gauss elimination b) Gauss-Jordan elimination 11) Solve the following equation to find the value of C1 using Crout method

C 1 +C 2+ C 3=7 ,C 1+ 2C 2+ 3C 3=16 ,C 1+ 3C 2 +4 C 3=22 a) 2 b) 1 c) 8

d)5

12) Solve the following equation to find the value of C3 using Crout method

C1 −2C 2+ 3C 3=6 ,C 1−C 2+2 C 3=9 , 3 C1 +2 C2−C 3=16 a) 8 b) 11 c) -77

d)13

13) Solve the following equation to find the value of C3 & C2 using Crout method

2C 1 +3 C 2 +1 C3=−1, 5 C 1 + 1 C2 + 1 C 3 =−9 , 3C 1 +2C 2 +4 C 3=11 21 108 21 22 47 21 a) C2 = ,C 3= b) C2 = , C 3= c) C2 = ,C 3= 18 5 8 7 2 8 C 2 =3,C 3=15 14) The rule

A=LU

[

L11 0 0 L21 L22 0 L31 L32 L33

,L=

]

in

a) Doolittle method 15) In L.U. decomposition Crout method, the entry U12 =… a)1 b)2 c)3

[ ] 3 5 1 0 4 7 0 0 8

16) This matrix

b) Crout method d)4

is

a) upper triangular b) Lower triangular 17) The three basic row operation can be used in L.U. decomposition: a) True b) False

[−63 −41 ] 6 5 11 3 1 0 3 1 a¿[ b¿[ 4 0 ][ 5 1 ] −2 1 ][ 0 −2]

18) Find L.U. decomposition of

[ ] 1 2 3 2 4 5 1 3 4

19) The matrix

?

c¿

[15 41][ 31 −51 ]

doesn’t have an L.U. decomposition

a) True b) False 20) Which of the following matrix have the property of L.U. decomposition?

[ ]

3 2 a¿ 0 1

[

1 −3 7 c ¿ −2 6 1 0 3 −2

[ ]

0 1 b ¿ 3 2

]

ANSWER THE FOLLOWING QUESTION: 1) Find the value of K that make these equations to make it non zero

4 H 1 + 9 H 2 + H 3= 0 ,16 H 1 + 26 H 2 + KH 3=0 , 9 H 1−8 H 2−3 H 3= 0 2) Find the value of unknown variables:

N 1+ 2 N 2 +3 N 3=5 , 3 N 1−N 2 +2 N 3=8 , 4 N 1−6 N 2−4 N 3=−2 3) Solve the equations using L.U. decomposition method:

2 f 1 + 3 f 2 + 4 f 3=3 , 3 f 1 + 7 f 2+7 f 3= 4 , 4 f 1 + 7 f 2+ 9 f 3=6 4) Solve the following system using Crout’s method

2 E 1−4 E2+ E3=4 , 6 E1 + 2 E 2−2 E3=10 ,−2 E1 +6 E2−2 E3=−4

11

d)

12...