Question 94544
Alright here goes the solution for your question on solving equations by the method of gaussian elimination 

2x + y - 3z = 1 

3x - y + 4z = 6 

x + 2y - z = 9 

The given system can be written in the augmented form as: 


[2 1 -3 : 1]
[3 -1 4 : 6]
[1 2 -1 : 9] 

The operations performed are: 

We shall first interchange the first row with the third row. 

[1 2 -1 : 9]
[3 -1 4 : 6]
[2 1 -3 : 1] 

Next: R2 - 3R1 and R3 - 2R1 are the operations which follow 

[1 2 -1 : 9]
[0 -7 7 : -21]
[0 -3 -1 : -17] 

R3 - R2 is the next opeartion. 

[1 2 -1 : 9]
[0 -7 7 : -21]
[0 4 -8 : 4] 

Lets divide R2/7 and R3/4 

[1 2 -1 : 9]
[0 -1 1 : -3]
[0 1 -2 : 1] 


Now R3 + R2 gives us the final step 

[1 2 -1 : 9]
[0 -1 1 : -3]
[0 0 -1 : -2] 

From this the given system of equations reduces to: 


x + 2y - z = 9 
-y + z = -3 
-z = - 2 

Hence, the value of z can be directly found out. 

==> z = 2 

now apply back substituion. 

-y + z = -3 

-y + 2 = -3 

-y = -3 - 2 

-y = -5 

==> y = 5 

Using y and z in the first equation we find the value of x. 

x + 2y - z = 9 

x + 10 - 2 = 9 

x = 9 - 8 

x = 1 

Hence, the solution. 

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Now the second question. 


The given set of equations are: 


x - y + z = 17 
x + y - z = -11 
x - y - z = 9 

This in the augmented form can be written as: 

[1 -1 1 : 17]
[1 1 -1 : -11]
[1 -1 -1 : 9] 

Lets perform the firts row operation that is R2 - R1 and R3- R1 


[1 -1 1 : 17]
[0 2 -2 : -28]
[0 0 -2 : -8] 

Now we divide R2/2 and R3/2 


[1 -1 1 : 17]
[0 1 -1 : -14]
[0 0 -1 : -4] 

Thus the above augmented matrix can be written as: 


x - y + z = 17 
y - z = -14 
-z = -4 

This implies z = 4 

By back substitution, we find the values of y and x. 

y - 4 = -14 

y = -14 + 4 

==> y = -10 

Now we find the value of x. 

x - y + z = 17 

x + 10 + 4 = 17 

x + 14 = 17 

x = 17 - 14 

x = 3 

Hence, the solution. 

I hope all the stpes are quite clear.. 

You can get back to me for an online help in math at any time of yours. 

Regards 
Chitra