How do you solve using gaussian elimination or gauss-jordan elimination, #2x - y + 5z - t = 7#, #x + 2y - 3t = 6#, #3x - 4y + 10z + t = 8#?

Question

1422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-12T09:57:55

The solutions are #((x),(y),(z),(t))=((4-2z+t),(1+z+t),(z),(t))#

We perform the Gauss Jordan elimination with the augmented matrix

#((1,2,0,-3,:,6),(2,-1,5,-1,:,7),(3,-4,10,1,:,8))#

#R3larrR3-3R1#, #=>#, #((1,2,0,-3,:,6),(2,-1,5,-1,:,7),(0,-10,10,10,:,-10))#

#R2larrR2-2R1#, #=>#, #((1,2,0,-3,:,6),(0,-5,5,5,:,-5),(0,-10,10,10,:,-10))#

#R3larr(R3)/(-10)#, #=>#, #((1,2,0,-3,:,6),(0,-5,5,5,:,-5),(0,1,-1,-1,:,1))#

#R2larr(R2)/(-5)#, #=>#, #((1,2,0,-3,:,6),(0,1,-1,-1,:,1),(0,1,-1,-1,:,1))#

#R3larrR3-R2#, #=>#, #((1,2,0,-3,:,6),(0,1,-1,-1,:,1),(0,0,0,0,:,0))#

#R1larrR1-2R2#, #=>#, #((1,0,2,-1,:,4),(0,1,-1,-1,:,1),(0,0,0,0,:,0))#

The solutions are

#((x),(y),(z),(t))=((4-2z+t),(1+z+t),(z),(t))#

#z# and #t# are free