How do you solve #5x+2y-z=-7#, #x-2y+2z=0#, and #3y+ z=17#?

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Answer 1 · 2017-08-04T14:50:54

The solution is #((x),(y),(z))=((-2),(4),(5))#

We perform the Gauss Jordan elimination with the augmented matrix

#((1,-2,2,:,0),(5,2,-1,:,-7),(0,3,1,:,17))#

#R2larrR2-5R1#

#((1,-2,2,:,0),(0,12,-11,:,-7),(0,3,1,:,17))#

#R3larr4R3-R2#

#((1,-2,2,:,0),(0,12,-11,:,-7),(0,0,15,:,75))#

#R3larrR3/15#

#((1,-2,2,:,0),(0,12,-11,:,-7),(0,0,1,:,5))#

#R2larrR2+11R3#

#((1,-2,2,:,0),(0,12,0,:,48),(0,0,1,:,5))#

#R2larrR2/12#

#((1,-2,2,:,0),(0,1,0,:,4),(0,0,1,:,5))#

#R1larrR1-2R3#

#((1,-2,0,:,-10),(0,1,0,:,4),(0,0,1,:,5))#

#R1larrR1+2R2#

#((1,0,0,:,-2),(0,1,0,:,4),(0,0,1,:,5))#