A New Formulae Method For Solving The Simultaneous Equations

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77

A New Formulae Method For Solving The

Simultaneous Equations

Avinash A. Musale, Eknath S. Ugale

ABSTRACT: Actually there are too many Methods which are used for solving the various types of equations by various methods. The equations may be like linear, quadratic, simultaneous, radial, and exponential as well as the equations with the number of variables. And for the Simultaneous Equation the methods are like Addition, Substitution, Elimination as well as Graphical Method also. We know there are too many methods like Gauss Elimination, Gauss Seidal, and Jacobi used for solving the Simultaneous Equations. But here we are going to introduce a new Method for solving the Simultaneous Equations as well as the comparison between the Methods above told and the new one. The paper will show you a new easy method for solving the Simultaneous Equation and the difference between the methods that are being used in the colleges which are Gauss Elimination, Gauss Seidal etc. Keywords: Simultaneous, Equations, Formulae, Method, Gauss, Elimination, Derivation

————————————————————

1. I

NTRODUCTION

:

The paper represents a method for solving the simultaneous equations in a different way. We use the substitution, addition, matrices Methods for solving the simultaneous equations but in this paper the method is related to directly the formulas for solving the simultaneous equations.

2. M

ETHOD

(F

ORMULAE

):

For Two Unknowns – X & Y

a11X+a12Y=b1; a21X+a22Y=b2; Solution:

u= [(a11*b2 - a21*b1)] v= [(a11*a22 - a12*a21)]

Y=

X=

For Three Unknowns – X, Y & Z

a11X+a12Y+a13Z=b1; a21X+a22Y+a23Z=b2; a31X+a32Y+a33Z=b3; Solution:

r= [(a11*b3) - (a31*b1)]; s= [(a11*b2) - (a21*b1)]; t= [(a11.a22) - (a12*a21)]; u= [(a11*a33) - (a13*a31)]; v= [(a11*a23) - (a13*a21)]; w=[(a11*a32) - (a12*a31)];

Z=* +

Y=* +

X=* +

For Four Unknowns – X, Y, Z & W

a11X+a12Y+a13Z+a14W=b1; a21X+a22Y+a23Z+a24W=b2; a31X+a32Y+a33Z+a34W=b3; a41X+a42Y+a43Z+a44W=b4; Solution:

k=[(a11*b4) - (a41*b1)]; l= [(a11*b3) - (a31*b1)]; m= [(a11*b2) - (a21*b1)]; n= [(a11.a22) - (a12*a21)]; o= [(a11*a33) - (a13*a31)]; p=[(a11*a44) - (a14*a41)]; q= [(a11*a23) - (a13*a21)]; r=[(a11*a24) - (a21*a14)]; s=[(a11*a32) - (a12*a31)]; t=[(a11*a34) - (a31*a14)]; u=[(a11*a42) - (a41*a12)]; v=[(a11*a43) - (a41*a13)];

g=(on-sq) h=(vn-uq) i=(ln-sm) j=(tn-sr) e=(kn-um)

f=(pn-ur)

W=* +

Z=* +

Y=* +

X=* + ___________________________

 Avinash A. Musale is currently pursuing bachelor degree in Mechanical Engineering in MIT AOE Pune in Savitribai Phule Pune University, India, Mobile No. -+918484972706. E-mail: [email protected]  Mr. Eknath S. Ugale is an Assistant Professor in the

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3. D

ERIVATION

:

For Two Unknowns – X & Y

a11X+a12Y=b1; a21X+a22Y=b2;

Using Gauss Elimination Method

Convert the given equations in matrix AX=B form

*𝑎11 𝑎12_{𝑎21 𝑎22}+ * *𝑋_𝑌+ = *𝑏1_𝑏2+

Make the Matrix A as Upper Triangular Matrix

Make row 1 as following Instruction R1 = R1/a11

* 1 𝑎12′

𝑎21 𝑎22+ * *𝑋𝑌+ = *𝑏1′𝑏2+

a12’=a12/a11 ……….(1)

b1’=b1/a11 …………...(2)

Make row 2 as following Instruction R2 = R2 - a21*R1

*1 𝑎12′

0 𝑎22′+ * *𝑋𝑌+ = *𝑏1′𝑏2′+

a22’=a22 - a21*a12’ ………… (3)

b2’=b2 – a21*b1’ ………(4)

Make row 2 as following Instruction R2 = R2/a22’

*1 𝑎12′

0 1 + * *𝑋𝑌+ = * 𝑏1′𝑏2′′+

b2’’=b2’/a22’ ………(5)

We Get….

*1 𝑎12′₀ ₁ + * *𝑋_𝑌+ = * 𝑏1′_𝑏2′′+

From the above solved Matrix

By the Backward Substitution we will get

Here Y=b2’’ ……….(a)

X+a12’Y=b1’ ………(b)

And from equation (a) Y=b2’’ From eq(5)

Y=* +

From equation (3),(4) Y=* +

From equation (1),(2) Y=* +

Y=**

+ * ++

a11 gets cancelled…..

Y=* +

and from given equation

or from equation (b) We get

X=* + or X=* +

If

u=[(a11*b2 - a21*b1)] v=[(a11*a22 - a12*a21)]

Y=

X=

For Three Unknowns – X, Y & Z

a11X+a12Y+a13Z=b1; a21X+a22Y+a23Z=b2; a31X+a32Y+a33Z=b3;

By Gauss Elimination Method

Convert the given equations in matrix AX=B form

[𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33

] * [𝑋𝑌 𝑍

] = [𝑏1𝑏2 𝑏3 ]

Make the Matrix A as Upper Triangular Matrix therefore Make row 1 as following Instruction

R1 = R1/a11

[𝑎21 𝑎221 𝑎12′ 𝑎13′𝑎23 𝑎31 𝑎32 𝑎33

] * [𝑋𝑌 𝑍

] = [𝑏1′𝑏2 𝑏3 ]

a12’= a12/a11 ………… (1)

a13’=a13/a11 …………(2)

b1’=b1/a11 ………….(3)

Make row 2 and row 3 as following Instruction R2 = R2 – a21* R1

R3 = R3-a31 *R1

[1 𝑎12′ 𝑎13′0 𝑎22′ 𝑎23′ 0 𝑎32′ 𝑎33′

] * [𝑋𝑌 𝑍]

= [𝑏1′𝑏2′ 𝑏3′ ]

a22’= a22 - a21*a12’ ……….(4)

a23’= a23 - a21*a13’ ……….(5)

b2’ = b2 – a21 * b1’ ……….(6)

a32’ = a32 – a31 * a12’ ……….(7)

a33’ = a33 – a31 * a13’ ……….(8)

b3’ = b3 – a31’ * b1’ ………..(9)

Make row 2 as following Instruction R2 = R2/a22

[1 𝑎12′ 𝑎13′0 1 𝑎23′′ 0 𝑎32′ 𝑎33′

] * [𝑋𝑌 𝑍] = [

𝑏1′ 𝑏2′′ 𝑏3′ ]

a23’’ = a23’/a22’ ………..(10)

b2’’ =b2’/a22’ ………..(11)

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R3 = R3 – a32’ * R2

[1 𝑎12′ 𝑎13′0 1 𝑎23′′ 0 0 𝑎33′′

] * [𝑋𝑌 𝑍]

= [𝑏2′′𝑏1′ 𝑏3′′ ]

a33’’ = a33’ - a32’ * a23’’ …………(12)

b3’’ = b3’ – a32’ * b2’’ …………(13)

Make row 3 as following Instruction R3 = R3/a33’’

[1 𝑎12′ 𝑎13′0 1 𝑎23′′

0 0 1

] * [𝑋𝑌 𝑍

] = [𝑏2′′𝑏1′ 𝑏3′′′ ]

b3’’’ = b3’’/a33’’ …………..(14)

From the above Matrix,

By the Backward Substitution we get

Z = b3’’’ ……..(a)

Y+a23’’Z=b2’’ ……..(b)

From equation (a) We get

Z = b3’’’

Z = * + …….from eq(14)

Z = * + …….fromeq(12)&(13)

Z = * +…..from eq(7,8,9,10,11)

Z = [*( (

))+ ( ( )) [

( ) ( )]

* ( )+ *( ( ))+ [( )_{( )}]]

……. From eq (1,2,3,4,5,6)

Z =

[

*(( ))+ *((

))+ [

( (₎₎

( ( ₎₎ ]

*( )+ *(( ))+ [

( ( ₎₎

( ( ₎₎ ]

]

…..From eq(1,2,3)

Z =

[

()*(( ))+ *(( ))+ [

(( ))

(( )) ]

()*( )+ *(( ))+ [

(( ))

(( )) ]

]

*+ gets cancelled

as well as a11 gets cancelled

Z =

[

()*(( ))+ *(( ))+ [

(( ))

(( )) ]

()*( )+ *(( ))+ [

(( ))

(( )) ]

]

Z =

[

*(( ))+ *(( ))+ [

(( ))

(( )) ]

*( )+ *(( ))+ [

(( ))

(( )) ]

]

Z = * [ ] [ ] * + [ ] [ ] * ++

Z=

* *

[ ] [ ] [ ]

+

*[ ] [ ] ++

𝑎11 𝑎22 − 𝑎21 𝑎12 gets cancelled… Z=

[ *

[ ] [ ] [ ]

+

*[ ] [ ] +]

Z=

*_{[ ] [ ]}[ ] [ ] [ ] +

If

r= [(a11*b3) - (a31*b1)]; s= [(a11*b2) - (a21*b1)]; t= [(a11.a22) - (a12*a21)]; u= [(a11*a33) - (a13*a31)]; v= [(a11*a23) - (a13*a21)]; w=[(a11*a32) - (a12*a31)];

Z=* +

From equation (b) Y+a23’’Z=b2’’

Y=[(b2’’)-(a23’’*Z)]

Y= * + - * +from eq(10,11)

Y=* +

Y=* ( )+

Y=* + …..from eq(4,5,6)

Y=[( (

)) (( ( )) )

( ( ))

]…..from eq(1,2,3)

Y=[((

)) ((( )) )

(( )) ]

Y=[( ) [((

_{)) (((} _{)) ])}

() (( )) ]

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Y=[( ) [((

_{)) (((} _{)) ])}

( ) ((

₎₎ ]

Y=* ( ) +

s= [(a11*b2) - (a21*b1)]; t= [(a11.a22) - (a12*a21)]; v= [(a11*a23) - (a13*a21)];

Y=* +

And from given equations we get X=* +

And here Simultaneously we can derive the formulas for the four unknowns in the same way above derived. As well as we can also derive the formulas for the Simultaneous equations with the number of derivatives.

4. COMPARISION

A. Gauss Elimination Method

x-3y+z=4 2x-8y+8z=-2 -6x+3y-15z=9

Convert into AX=B Matrix form

Make the Matrix A as Upper Triangular Matrix

[12 −3−8 18 −6 3 −15] * [

𝑋 𝑌 𝑍] = [

4 −2

9]

Make row 2 and row 3 as following Instruction R2 = R2 – 2* R1

R3 = R3-(-6) *R1

[10 −3−2 16 0 −15 −9] * [

𝑋 𝑌 𝑍] = [

4 −10

33 ]

Make row 2as following Instruction R2 = R2/ (-2)

[10 −31 −31 0 −15 −9] * [

𝑋 𝑌 𝑍] = [

4 5 33]

Make row 3 as following Instruction R3 = R3 – (-15) * R2

[1 −30 1 −31 0 0 −54]

* [𝑋𝑌 𝑍]

= [ 45 108]

And then by Backward Substitution Method We will get

-54*Z=108 Z=-2

Y-3*Z=5 Y=-1

And finally

X=3

Hence we can conclude

X=3 Y=-1 Z=-2

B. New Formula Method

x-3y+z=4 2x-8y+8z=-2 -6x+3y-15z=9

Compare it with a11X+a12Y+a13Z=b1; a21X+a22Y+a23Z=b2; a31X+a32Y+a33Z=b3;

Then here for three unknowns Substitute the values in the given formulae Hence we will get

r= [(a11*b3) - (a31*b1)]; r= 33

s= [(a11*b2) - (a21*b1)]; s= -10

t= [(a11.a22) - (a12*a21)]; t= -2

u= [(a11*a33) - (a13*a31)]; u= -9

v= [(a11*a23) - (a13*a21)]; v=6

w=[(a11*a32) - (a12*a31)]; w=-15

Z=* + = *[ ] [ ]_{[ ] [ ]} +

Z= -2

Y=* + = * [ ] +

Y= -1

X=* += * +

X=3

Hence X=3, Y= -1, Z= -2

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5. CONCLUSION:

Here we can conclude that there is no confusion in solving the simultaneous equation by the Formulae Method. As well as it is easier and takes less time than the other methods like Gauss Elimination Method, Gauss Seidal Method, etc. And in the Engineering Calculators like CASIO fx-991ES PLUS there is no solution for the simultaneous equation with four unknowns so we can use these formulas for solving the Simultaneous equations easily as well as fast.

6. REFERENCES:

[1] Steven C. Chapra, Raymond P. Canale, Numerical Method for Engineers. 4/e, Tata McGraw Hill Editions, 2002, ISBN 0-07-04743-0

[2] Dr. B. S. Garewal, Numerical Method in Engineering and Science, 7/e, Khanna Publishers, ISBN81-74009-205-6

[3] Steven c. Chapra, Applied Methods in MATLAB for Engineers and Scientist, 2/e, Tata McGrawHill Publishing Co.Ltd, 2008, ISBN0-07-064853-0

[4] M. K. Jain, S R K Iyengar, R K Jain, Numerical Method for Scientific and Engineering Computations, 5/e New Age International Publisher ISBN 13 978-81-224-2001-2008

[5] Gary D. Knott, "Gaussian Elimination and LU-Decomposition", www.civilized.com, Feb 28, 2012

[6] Joseph F. Grcar, ―Mathematicians of Gaussian Elimination‖, Notices of the AMS, June 2011