SOLUTION: How do you solve a linear equations with Gaussian elimination? x+ y + Z =9 2X- 3y +4z =7 X- 4y + 3Z =-2

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: How do you solve a linear equations with Gaussian elimination? x+ y + Z =9 2X- 3y +4z =7 X- 4y + 3Z =-2      Log On


   



Question 206096: How do you solve a linear equations with Gaussian elimination?
x+ y + Z =9
2X- 3y +4z =7
X- 4y + 3Z =-2

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28matrix%283%2C4%2C1%2C1%2C1%2C9%2C2%2C-3%2C4%2C7%2C1%2C-4%2C3%2C-2%29%29

Multiply row 1 thru by -2 and add to row 2

%28matrix%283%2C4%2C1%2C1%2C1%2C9%2C0%2C-5%2C2%2C-11%2C1%2C-4%2C3%2C-2%29%29

Multiply row 1 by -1 and add to row 3

%28matrix%283%2C4%2C1%2C1%2C1%2C9%2C0%2C-5%2C2%2C-11%2C0%2C-5%2C2%2C-11%29%29

Multiply row 2 by -1 and add to row 3

%28matrix%283%2C4%2C1%2C1%2C1%2C9%2C0%2C-5%2C2%2C-11%2C0%2C0%2C0%2C0%29%29

Now we write the system of equations that this
matrix represents:



The bottom equation is all 0's so any values of
x, y, and z would be solutions to it, so we eliminate
it

Choose an arbitrary value for z, we will use z=a

Or, after erasing the 0s and 1 coefficients:

system%28x+%2B+y+%2B+z+=+++9%2C%0D%0A+-5y+%2B+2z+=+-11%29

Substitute a for z in the 2nd equation:

-5y+%2B+2a+=+-11

Solve for y

-5y=-11-2a

y=11%2F5%2B2a%2F5

Substitute a for z and 11%2F5%2B2a%2F5 
for y in the first equation:

x+%2B+y+%2B+z+=+++9

x+%2B+%2811%2F5+%2B2a%2F5%29+%2B+a+=+9

x+%2B+11%2F5+%2B+2a%2F5+%2B+a+=+9

Clear of fractions:

5x+%2B+11+%2B+2a+%2B+5a+=+45

5x+%2B+11+%2B+7a+=+45

5x+=+34+-+7a

+x+=+34%2F5+-+7a%2F5

So the solution is:

(x, y, z) = (34%2F5-7a%2F5,11%2F5%2B2a%2F5,a) 

Edwin