How do you solve using gaussian elimination or gauss-jordan elimination, #x_1 +2x_2 − x_3 +3x_4 =2#, #2x_1 + x_2 + x_3 +3x_4 =1#, #3x_1 +5x_2 − 2x_3 +7x_4 =3#, #2x_1 +6x_2 − 4x_3 +9x_4 =8#?

1 Answer
Aug 7, 2017

Answer:

The solutions are #x_4=2 #, # x_3# free, #x_1=-2-x_3, x_2=-1+x_3#

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

#((1,2,-1,3,:,2),(2,1,1,3,:,1),(3,5,-2,7,:,3),(2,6,-4,9,:,8))#

#R4larrR4-2R1#, #=># , #((1,2,-1,3,:,2),(2,1,1,3,:,1),(3,5,-2,7,:,3),(0,2,-2,3,:,4))#

#R3larrR3-3R1#, #=># , #((1,2,-1,3,:,2),(2,1,1,3,:,1),(0,-1,1,-2,:,-3),(0,2,-2,3,:,4))#

#R2larrR2-2R1#, #=># , #((1,2,-1,3,:,2),(0,-3,3,-3,:,-3),(0,-1,1,-2,:,-3),(0,2,-2,3,:,4))#

#R2larrR1-R4#, #=># , #((1,0,1,0,:,-2),(0,-3,3,-3,:,-3),(0,-1,1,-2,:,-3),(0,2,-2,3,:,4))#

#R4larrR4+2R3#, #=># , #((1,0,1,0,:,-2),(0,-3,3,-3,:,-3),(0,-1,1,-2,:,-3),(0,0,0,-1,:,-2))#

#R4larr(R4)/(-1)#, #=># , #((1,0,1,0,:,-2),(0,-3,3,-3,:,-3),(0,-1,1,-2,:,-3),(0,0,0,1,:,2))#

#R2larr(R2)/(-3)#, #=># , #((1,0,1,0,:,-2),(0,1,-1,1,:,1),(0,-1,1,-2,:,-3),(0,0,0,1,:,2))#

#R3larrR3+R2#, #=># , #((1,0,1,0,:,-2),(0,1,-1,1,:,1),(0,0,0,-1,:,-2),(0,0,0,1,:,2))#

#R3larr(R3)/(-1)#, #=># , #((1,0,1,0,:,-2),(0,1,-1,1,:,1),(0,0,0,1,:,2),(0,0,0,1,:,2))#

#R4larrR4-R3#, #=># , #((1,0,1,0,:,-2),(0,1,-1,1,:,1),(0,0,0,1,:,2),(0,0,0,0,:,0))#

#R2larrR2-R3#, #=># , #((1,0,1,0,:,-2),(0,1,-1,0,:,-1),(0,0,0,1,:,2),(0,0,0,0,:,0))#

The solutions are #x_4=2 #, # x_3# free, #x_1=-2-x_3, x_2=-1+x_3#