Gauss elimination method

Gauss elimination method :

Introduction:

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Carl Friedrich Gauss[/caption]

Many times we are required to find out solution of linear equations. We also know that, we can find out roots of linear equations if we have sufficient number of equations. For example if we have to calculate three unknown variables, then we must have three equations. Many times we have solved such problems by eliminating one of the root and keep on decreasing number of variables. But in some cases it is not possible or it will take more time to solve.

Gauss elimination method is one of the simple and famous methods used for finding roots of linear equations. Let us discuss this method assuming we have three linear equations in x, y and z. That is we have to find out roots of that equations (values of x, y and z).

Steps to find out roots of linear equations using Gauss elimination method:

In this method we just eliminate ‘x’ from first equation using second and third equation. After that we get only two equations with two unknowns. Similarly, we then eliminate ‘y’ from first (among two equations that we get from last step) equation using second equation. Finally we get single equation in z having constant in its right side. Now we can find ‘z’, using ‘z’ we can find ‘y’, similarly ‘x’.

Don’t get confused, I will explain each step clearly.

Let us consider three linear equations as follows:

a1x+b1y+c1z=d1…………………1)

a2x+b2y+c2z=d2…………………2)

a3x+b3y+c3z=d3…………………3)

From above three equations we are asked for finding values of x, y and z (values of a1, b1, c1,……..,d3 are given).

Step 1: Eliminate ‘x’ from first equation using second and third equation. For doing this we have to subtract 1^st eq. from 2^nd eq. by making coefficient of ‘x’ (of 1^st equation) equals to coefficient of ‘x’ (of 2^nd equation). Similarly we have to do same thing for third equation.

In short we have to solve following equations:

eq.(2) – (a2/a1)*eq.(1) and eq.(3) – (a3/a1)*eq.(1)

We get two new equations in y and z as follows:

b2’y+c2’z=d2’……..4)

b3’y+c3’z=d3’……..5)

Step 2: similarly, we have to eliminate ‘y’ from 4^th equation using 5^th equation.

We have to solve following equation.

eq.(5) – (b3’/b2’)*eq.(4)

we get c3”z=d3”……………6)

solving above equation we get, z=d3”/c3”

Step 3: Finally we have to put above value of ‘z’ in equation 4) (or (5)), then we get ‘y’. now we have two roots (y and z). put ‘y’ and ‘z’ in eq.(1) (or (2) or eq.(3)), we will get ‘x’.

Example 1:

Find the roots of following equations using Gauss Elimination method.

X + 4y – z = -5 ………….1)

X + y - 6z = -12…………2)

3x – y – z = 4 ………3)

Solution:

Step 1: Perform eq.(2) – (a2/a1)*eq.(1) and eq.(3) – (a3/a1)*eq.(1)

We get, 3y +5z =7……………4)

And

-13y +2z = 19…………..5)

Step 2: Now perform eq.(5) – (b3’/b2’)*eq.(4)

We get, -13y + 2z – (-13/3)*(3y + 5z) = 19 + (13/3)*7

71z = 148

i.e. z=148/71.

Step 3: From eq.(5) - 13y = 19 - 2*(148/71)

= 19 – 296/71

Y= -81/71

From eq.(1) x+4(-81/71)-148/71=-5

Therefore, x=117/71.

Thus roots of given linear equations using Gauss elimination method is

X=117/71; y=-81/71; z=148/71.

Example 2:

Find roots of following linear equations using Gauss Elimination method:

2x+y+z=10………1)

3x+2y+3z=18……..2)

X+4y+9z=16………..3)

Solution :

Step 1: Perform eq.(2)-(3/2)*eq(1) and eq(3)-(1/2)*eq(1)

We get, y+3z=6…………4)

And 7y+17z=22……….5)

Step 2: Perform eq(5) - 7*eq(4)

We get, -4z=-20

i.e. z=5;

Step 3: From equation (5) y=-63/7 i.e. y=-9

From equation (1) 2x=14 i.e. x=7

Thus the roots of given linear equations is

X=7; y=-9; z=5.

Gauss elimination method