Problem13:Find the conditions that b's must satisfy for the system to be consistent 
> ;
x  2 y  z = b1 4 x + 5 y + 2 z = b2 4 x + 7 y + 4 z = b3

Perform Gaussian elimination on to the augmented matrix for the system and then read off the conditions from the final matrix in rref. 
> A := matrix(3,4,[1,2,1,b1, 4,5,2,b2, 4,7,4,b3]);
[ 1 2 1 b1] [ ] A := [4 5 2 b2] [ ] [4 7 4 b3]
there many ways to bring A into reduced row echelon form (e.g. using rref(A) will do). Here we use the function pivot.... 
> pivot(A,1,1);
[1 2 1 b1 ] [ ] [0 3 2 4 b1 + b2] [ ] [0 1 0 4 b1 + b3]> pivot(%,2,2);
[1 0 1/3  5/3 b1  2/3 b2 ] [ ] [0 3 2 4 b1 + b2 ] [ ] [0 0 2/3 8/3 b1  1/3 b2 + b3]> pivot(%,3,3);
[1 0 0 3 b1  1/2 b2  1/2 b3] [ ] [0 3 0 12 b1 + 3 b3 ] [ ] [0 0 2/3 8/3 b1  1/3 b2 + b3 ]
finally let's get leading ones with, 
> mulrow(%,2,1/3);
[1 0 0 3 b1  1/2 b2  1/2 b3] [ ] [0 1 0 4 b1  b3 ] [ ] [0 0 2/3 8/3 b1  1/3 b2 + b3 ]> mulrow(%,3,3/2);
[1 0 0 3 b1  1/2 b2  1/2 b3] [ ] [0 1 0 4 b1  b3 ] [ ] [0 0 1 4 b1  1/2 b2 + 3/2 b3 ]
Notice that the same thing could have been obtained with, 
> rref(A);
[1 0 0 3 b1  1/2 b2  1/2 b3] [ ] [0 1 0 4 b1  b3 ] [ ] [0 0 1 4 b1  1/2 b2 + 3/2 b3 ]
the left hand side is the 3x3 identity so the system has unique solution for every choice of b's. In other words the system is consistent no matter what b's we use for the right hand side. 