How do you solve using gaussian elimination or gauss-jordan elimination, #y + 3z = 6#, #x + 2y + 4z = 9#, #2x + y + 6z = 11#?

1 Answer
Apr 1, 2016

Answer:

Using Gaussian elimination
#color(white)("XXX")(x,y,z)= (1/7,9/7,11/7)#

Explanation:

Converting the given equations to augmented matrix (to avoid having to re-write the variable names)
#{(0,1,3,6),(1,2,4,9),(2,1,6,11):}color(white)("XX"){:([1]),([2]),([3]):}#

Exchanging rows [1] and [2]
#{(1,2,4,9),(0,1,3,6),(2,1,6,11):}color(white)("XX"){:([4]),([5]),([3]):}#

Note that at this point rows [4] and [5] are in Gaussian form
and all that remains is to zero out the first two columns of row [3]

Subtracting #2# times row [4] from row [3]
#{(1,2,4,9),(0,1,3,6),(0,-3,-2,-7):}color(white)("XX"){:([4]),([5]),([6]):}#

Adding #3# times row [5] to row [6]
#{(1,2,4,9),(0,1,3,6),(0,0,7,11):}color(white)("XX"){:([4]),([5]),([7]):}#

Dividing row [7] by #7#
#{(1,2,4,9),(0,1,3,6),(0,0,1,11/7):}color(white)("XX"){:([4]),([5]),([8]):}#

This gives us
#color(white)("XXX")z=11/7#

Substituting back into row [5]
#color(white)("XXX")y+3xx11/7=6 rarr y= 9/7#

Substituting back into row [4]
#color(white)("XXX")x+2xx9/7+4xx11/7=9 rarr x=1/7#