SOLUTION: How can i solve this using Gaussian elimination method? x+y+z=4 2x-y+3z=4 4x+2y-z=-15

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: How can i solve this using Gaussian elimination method? x+y+z=4 2x-y+3z=4 4x+2y-z=-15      Log On


   



Question 361018: How can i solve this using Gaussian elimination method?
x+y+z=4
2x-y+3z=4
4x+2y-z=-15

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
How can I solve this using Gaussian elimination method?
system%28x+%2B+y+%2B+z=+4%2C%0D%0A2x+-+y+%2B+3z=+4%2C%0D%0A4x+%2B+2y+-+z=+-15%29

Write that as a matrix by dropping the letters
and putting vertical line instead of equal signs:



The idea is to get three zeros in the three positions
in the lower left corner of the matrix, where the elements
I've colored red are:

To get a 0 where the red 2 on the left of the middle row is,
multiply R1 by -2 and add it to 1 times R2, and put it in place 
of the present R2.  That's written as

-2R1+1R2->R2

To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R1 and a 1 by R2

matrix%283%2C1%2C-2%2C1%2C%22%22%29


We are going to change only R2.  Although R1 gets multiplied
by -2 we are going to just do that mentally and add it to R2, but
not really change R1.



-----

To get a 0 where the lower left red 4 is, multiply R1
by -4 and add it to 1 times R3.  That's written as

-4R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -4 by R1 and a 1 by R3

matrix%283%2C1%2C-4%2C%22%22%2C1%29


We are going to change only R3. 




---------------

To get a 0 where the red -2 is, multiply R2
by -2 and add it to 3 times R3.  That's written as

-2R2+3R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R2 and a 3 by R3

matrix%283%2C1%2C%22%22%2C%22-2%22%2C3%29

We are going to change only R3. 



Now that we have 0's in the three positions in the
lower left corner of the matrix, we change the matrix
back to equations:

system%28x%2By%2Bz=4%2C-3y%2Bz=-4%2C-17z=-85%29

Solve the third equation for z:

-17z=-85
z=%28-85%29%2F%28-17%29
z=5

Substitute 5 for z in the middle equation:

-3y%2Bz=-4
-3y%2B%285%29=-4
-3y%2B5=-4
-3y=-9
y=3

Substitute 5 for z and 3 for y in the top equation:

x%2By%2Bz=4
x%2B%283%29%2B%285%29=4
x%2B3%2B5=4
x%2B8=4
x=-4

So the solution is %22%28x%2Cy%2Cz%29%22=%22%28-4%2C3%2C5%29%22

Edwin