How do you solve using gaussian elimination or gauss-jordan elimination, #2x + y - 3z = - 3#, #3x + 2y + 4z = 5#, #-4x - y + 2z = 4#?

1 Answer
Jul 3, 2018

Answer:

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

Explanation:

Perform the Gauss-Jordan Elimination on the augmented matrix

#A=((2,1,-3,|,-3),(3,2,4,|,5),(-4,-1,2,|,4))#

Make the pivot in the first column #R1larr(R1)/2#

#((1,1/2,-3/2,|,-3/2),(3,2,4,|,5),(-4,-1,2,|,4))#

Eliminate the first column #R2larr(R2-3R1)# and #R3larr(R3+4R1)#

#((1,1/2,-3/2,|,-3/2),(0,1/2,17/2,|,19/2),(0,1,-4,|,-2))#

Make the pivot in the #2#nd column swap #R2harrR3#

#((1,1/2,-3/2,|,-3/2),(0,1,-4,|,-2),(0,1/2,17/2,|,19/2))#

Eliminate the second column #R3larr(R3-1/2(R2))#, #R1larr(R1-(R2)/2)#

#((1,0,1/2,|,-1/2),(0,1,-4,|,-2),(0,0,21/2,|,21/2))#

Make the pivot in the #3rd# column #R3larr(R3*2/21)#

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

Eliminate the #3rd # column #R1larr(R1-(R3)/2)# and #R2larr(R2+4R3)#

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

The solution is

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