How do you solve #[(-x,+2y,-4z,-5u,=,27),(-2x,-3y,+3z,-3u,=,9),(5x,-4y,+2z,-3u,=,5), (-2x,+0y,+4z,-5u,=,29)]#?

1 Answer
Mar 5, 2018

Answer:

#x=0#, #y=3#, #z=1# and #u=-5#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((-1,2,-4,-5,|,27),(-2,-3,3,-3,|,9),(5,-4,2,-3,|,5),(-2, 0, 4, -5,|,29))#

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

#R2larrR2-2R1#, #R3larrR3+5R1, R4larrR4-2R1#;

#A=((-1,2,-4,-5,|,27),(0,-7,11,7,|,-45),(0,6,-18,-28,|,140),(0, -4, 12, 5,|,-25))#

#R2larrR2-2R4#;

#A=((-1,2,-4,-5,|,27),(0,1,-13,-3,|,5),(0,6,-18,-28,|,140),(0, -4, 12, 5,|,-25))#

#R1larrR1-2R2#, #R3larrR3-6R2, R4larrR4+4R2#;

#A=((-1,0,22,1,|,17),(0,1,-13,-3,|,5),(0,0,60,-10,|,110),(0, 0, -40, -7,|,-5))#

#R1larrR1#, #R3larrR3*2#, #R4larrR4*3#;

#A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, -120, -21,|,-15))#

#R4larrR4+R3#;

#A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, 0, -41,|,205))#

#R4larr(R4)/(-41)#;

#A=((1,0,-22,-1,|,-17),(0,1,-13,-3,|,5),(0,0,120,-20,|,220),(0, 0, 0, 1,|,-5))#

#R1larrR1+R4#, #R2larrR2+3R4#, #R3larrR3+20R4#;

#A=((1,0,-22,0,|,-22),(0,1,-13,0,|,-10),(0,0,120,0,|,120),(0, 0, 0, 1,|,-5))#

#R3larr(R3)/120#;

#A=((1,0,-22,0,|,-22),(0,1,-13,0,|,-10),(0,0,1,0,|,1),(0, 0, 0, 1,|,-5))#

#R1larrR1+22R3#, #R2larrR2+13R3#;

#A=((1,0,0,0,|,0),(0,1,0,0,|,3),(0,0,1,0,|,1),(0, 0, 0, 1,|,-5))#

Thus #x=0#, #y=3#, #z=1# and #u=-5#