How do you solve using gaussian elimination or gauss-jordan elimination, #x+y+z=1#, #x+y-2z=3#, #x+2y+z=2#?

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Answer 1 · 2018-07-11T14:36:27

The solution is #((x),(y),(y))=((2/3),(1),(-2/3))#

Perform the Jordan-Gauss elimination on the augmented matrix

#A=((1,1,1,|,1),(1,1,-2,|,3),(1,2,1,|,2))#

The pivot is in the first colum and the first row.

Eliminate the first column by performing the row operations

#R2larrR2-R1# and #R3larrR3-R1#

#((1,1,1,|,1),(0,0,-3,|,2),(0,1,0,|,1))#

Swap the last two rows

#R2harrR3#

#((1,1,1,|,1),(0,1,0,|,1),(0,0,-3,|,2))#

The pivot is in the second row and second column

Eliminate the second column

#R1larrR1-R2#

#((1,0,1,|,0),(0,1,0,|,1),(0,0,-3,|,2))#

Make the pivot in the third row of third column

#R3larrR3*(-1/3)#

#((1,0,1,|,0),(0,1,0,|,1),(0,0,1,|,-2/3))#

Eliminate the third column

#R1larrR1-R3#

#((1,0,0,|,2/3),(0,1,0,|,1),(0,0,1,|,-2/3))#

The solution is

#((x),(y),(y))=((2/3),(1),(-2/3))#