Thread: Gaussian elimination with back subbing

Hi, I need help with a question -

I need to use Gaussian elimination and back substitution to solve the following system:

0 + 4x2 – 3x3 = 3
-x1 + 7x2 – 5x3 = 4
-x1 + 8x2 - 6x3 = 5

Since the first pivotal position is 0, I interchange rows one and two before eliminating numbers below the pivot. I'm not sure what to do after that.


Thanks for any help.

Originally Posted by Kenny12345

Hi, I need help with a question -

I need to use Gaussian elimination and back substitution to solve the following system:

0 + 4x2 – 3x3 = 3
-x1 + 7x2 – 5x3 = 4
-x1 + 8x2 - 6x3 = 5

Since the first pivotal position is 0, I interchange rows one and two before eliminating numbers below the pivot. I'm not sure what to do after that.


Thanks for any help.

interchange first and last equation:

-x1 + 8x2 - 6x3 = 5
-x1 + 7x2 – 5x3 = 4
0 + 4x2 – 3x3 = 3

Subtract equation 1 from 2:

-x1 + 8x2 - 6x3 = 5
0 - 1x2 + x3 = -1
0 + 4x2 – 3x3 = 3

Add 4 times the second to the last:

-x1 + 8x2 - 6x3 = 5
0 - 1x2 + x3 = -1
0 + 0 + 1x3 = -1

Now start the back substitution with x3=-1, put this into the second to find x2, etc..

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The solution in my book said the answer is (1,0,-1) - it doesn't offer an explanation though. Also, why did you interchange the first and last equations instead of the first and second equations?

Originally Posted by Kenny12345

The solution in my book said the answer is (1,0,-1) - it doesn't offer an explanation though. Also, why did you interchange the first and last equations instead of the first and second equations?

Arithmetic error corrected.

.

Thanks for the help , but I still wonder why you would exchanged the 1st equation with the 3rd equation instead of exchanging the 1st and 2nd?