How do you solve using gaussian elimination or gauss-jordan elimination, #2x+4y-6z=48#, #x+2y+3z=-6#, #3x-4y+4z=-23#?

1 Answer
May 19, 2017

Answer:

Using Gauss elimination gives #x=3#, #y=3#, and #z=-5#

Explanation:

Step 1. Rewrite the equations in matrix form
#((2,4,-6,|,48),(1,2,3,|,-6),(3,-4,4,|,-23))#

Step 2. Switch #R_2# with #R_1#
#((1,2,3,|,-6),(2,4,-6,|,48),(3,-4,4,|,-23))#

Step 3. #-2R_1+R_2->R_2#
#((1,2,3,|,-6),(0,0,-12,|,60),(3,-4,4,|,-23))#

Step 4. #-3R_1+R_3->R_3#
#((1,2,3,|,-6),(0,0,-12,|,60),(0,-10,-5,|,-5))#

Step 5. Switch #R_2# with #R_3#
#((1,2,3,|,-6),(0,-10,-5,|,-5),(0,0,-12,|,60))#

Step 6. #-1/10R_2 -> R_2#
#((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,-12,|,60))#

Step 7. #-1/12R_3->R_3#
#((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,1,|,-5))#

Translating back into the values #x#, #y#, and #z#
#z=-5#
#y+1/2(-5)=1/2=>y=3#
#x+2(3)+3(-5)=-6=>x=3#