Establish the conditions under which the equationsax+by+cz=q−r;bx+cy+az=r−p;cx+ay+bz=p−q,are consistent. I am aware that by taking the system to echelon can get me the rank of the coefficient matrix and the augmented matrix. But that turns out to be a very cumbersome process and the terms in the echelon form are very big. Is there other ways to establish the conditions for consistency which are less cumbersome and time consuming?

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homofilirix · Accepted Answer

The right-hand sides of the three equations suggest that adding them all together might be fruitful. When you do this, you get, after a little bit of fiddling,  (a+b+c)(x+y+z)=0.  This equation can be satisfied for any values of a, b and c, so the original system is also consistent for all values of these parameters, as well as all values of p and q, since the latter don’t appear in this equation at all.  Another approach is to use Cramer’s rule, just as in an answer to your previous question.  det((a   b   c),(b   c   a),(c   a   b))=3abc−(a3+b3+c3)  This is equal to zero when a+b+c=0, so there are two cases to consider. If a+b+c≠0, there is a unique solution to this system of equations for any values of p and q. If it’s zero, you’ll need to look a little more closely to determine consistency.  One way to proceed in that case is to substitute c=−a−b in the augmented matrix of the system and row-reduce. The computation is a bit messy, but not as bad a row-reducing the original augmented matrix. The system is consistent (thought indeterminate) iff all of the zero rows on the left (unaugmented) side of the rref have zeros in the right-hand column. (You’ll end up with denominators of a2+ab+b2, but that can’t be zero.) Another way to go is to perform this substitution in the three equations, and then add them together as in the first part of this answer. You end up with 0=0, which is obviously consistent.

Establish the conditions under which the equations ax+by+cz=q-r;bx+cy+az=r-p;cx+ay+bz=p-qZS

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