How do you solve the system #x+y+z=1#, #2x-y+2z=-1#, and #-x-3y+z=1#?

2 Answers
Jul 17, 2018

Answer:

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

Explanation:

Perform the Gauss- Jordan Elimination on the Augmented Matrix

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

The pivot is in the first column and first row

Eliminate the first column

#R_2larr(R2-2R1)# and #R3larr(R3+R1)#

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

Make the pivot in the second row of the second column

#R2larr(R2)/(-3)#

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

Eliminate the second column

#R1larr(R1-R2)#

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

#R3larrR3+2R2#

#((1,0,1,|,0),(0,1,0,|,1),(0,0,2,|,4))#

Make the pivot in the third column

#R3larr(R3/2)#

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

Eliminate the third column

#R1larr(R1-R3)#

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

The solution is

#((x),(y),(z))=((-2),(1),(2))#

Answer:

#x=-2, y=1, z=2#

Explanation:

Given equations:

#x+y+z=1\ ...........(1)#

#2x-y+2z=-1\ ...........(2)#

#-x-3y+z=1\ ...........(3)#

Adding (1) & (2), we get

#x+y+z+2x-y+2z=1+(-1)#

#3x+3z=0#

#x+z=0\ ........(4)#

Multiplying (1) by #3# & adding to (3) as follows

#-x-3y+z+3(x+y+z)=1+3\cdot 1#

#2x+4z=4#

#x+2z=2\ .........(5)#

subtracting (4) from (5) as follows

#x+2z-(x+z)=2-0#

#z=2#

Setting #z=2# in (4), we get

#x+2=0#

#x=-2#

Now, setting #x=-2# & #z=2# in (1), we get

#-2+y+2=1#

#y=1#

Hence, we get

#x=-2, y=1, z=2#