SOLUTION: 3x-6y+z=0 x+y=0 -12x+24y-4z=0

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Question 378061: 3x-6y+z=0
x+y=0
-12x+24y-4z=0

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
  3x -  6y + 1z = 0
  1x +  1y + 0z = 0
-12x + 24y - 4z = 0

We make the augmented matrix by eliminating the variables:



Swap R1 and R2  (row 1 and row 2)
That instruction is written
R1<-->R2



Get a 0 where the 3 is by multiplying R1 by -3
temporarily and add it to row 2, then restore R1

That instruction is written

-3*R1+R2 -> R2



Get a 0 where the -12 is by multiplying R1 by 12
temporarily and add it to row 2, then restore R1

That instruction is written

12*R1+R3 --> R3



Get a 1 where the -9 is by multiplying row 2 by -1%2F9

That instruction is written:

-1%2F9R2 --> R2



Get a 0 where the 36 is by multiplying R2 by -36
temporarily and add it to row 3, then restore R2

That instruction is written

-36R2+R3-->R3



Write that as a system of equations:



or 

system%28++x+%2B++y+=+0%2C%0D%0A++y+-+expr%281%2F9%29z+=+0%29

We let z be arbitrary.  Solve the second equation for y

y=expr%281%2F9%29z

Solve the first equation for x:

x+=+-y

Substitute expr%281%2F9%29z for y

x+=+-expr%281%2F9%29z

So the set of solutions is

(x,y,z) = (-expr%281%2F9%29z,expr%281%2F9%29z,z)

where z is any arbitrary number.

Edwin