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
