How do you solve #-x + 2y - 3z = 0 #, #2x + z = 0 #, and #3x - 4y + 4z = 2# using matrices?

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Answer 1 · 2018-03-17T10:45:46

The solution is #S=((x=2/5),(y=-1),(z=-4/5))#

Perform the Gauss- Jordan elimination on the augmented matrix

#A=((-1,2,-3,|,0),(2,0,1,|,0),(3,-4,4,|,2))#

Make the pivot in the first column by

#R_1larr(-1)xxR_1#

#=((1,-2,3,|,0),(2,0,1,|,0),(3,-4,4,|,2))#

Eliminate the first column by

#R_2larrR_2-2R_1# and #R_3larrR_3-3R_1#

#=((1,-2,3,|,0),(0,4,-5,|,0),(0,2,-5,|,2))#

Make the pivot in the second column by

#R_2larrR_2/4#

#=((1,-2,3,|,0),(0,1,-5/4,|,0),(0,2,-5,|,2))#

Eliminate the second column by

#R_3larrR_3-2R_2# and #R_1larrR_1+2R_2#

#=((1,0,1/2,|,0),(0,1,-5/4,|,0),(0,0,-5/2,|,2))#

Make the pivot in the third column by

#R_3larrR_3/(-5/2)#

#=((1,0,1/2,|,0),(0,1,-5/4,|,0),(0,0,1,|,-4/5))#

Eliminate the third column by

#R_1larrR_1-1/2R_3# and #R_2larrR_2+5/4R_3#

#=((1,0,0,|,2/5),(0,1,0,|,-1),(0,0,1,|,-4/5))#

The solution is

#S=((x=2/5),(y=-1),(z=-4/5))#