How do you solve the system #4r-4x+4t=-4#, #4r+x-2t=5#, and #-3r-3x-4t=-16#?

1 Answer
Feb 23, 2018

Answer:

#r=1#, #x=3# and #t=1#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((4,-4,4,|,-4),(4,1,-2,|,5),(-3,-3,-4,|,-16))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R1larr(R1)/4#

#A=((1,-1,1,|,-1),(4,1,-2,|,5),(-3,-3,-4,|,-16))#

#R2larrR1-4R2#; #R3larrR3+3R1#

#A=((1,-1,1,|,-1),(0,5,-6,|,9),(0,-6,-1,|,-19))#

#R2larrR2*6#; #R3larrR3*5#

#A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,-30,-5,|,-95))#

#R3larrR3+R2#

#A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,0,-41,|,-41))#

#R3larr(R3)/(-41)#

#A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,0,1,|,1))#

#R1larrR1-R2; R2larrR2+36R3#

#A=((1,-1,0,|,-2),(0,30,0,|,90),(0,0,1,|,1))#

#R2larr(R2)/30#

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

#R1larrR1+R2#

#A=((1,0,0,|,1),(0,1,0,|,3),(0,0,1,|,1))#

Thus, #r=1#, #x=3# and #t=1#