Which condition $p$, $q$ and $r$ should satisfy so that the following system of linear simultaneous equations has at least one solution, such that $p+q+r\neq0?$
$x+2y-3z=p$
$2x+6y-11z=q\\x-2y+7z=r$
I approached this problem using Gauss Elimination method by creating an augmented matrix and reducing it to row echelon form. I got something like this :- $$ \left[\begin{array}{rrr|r} 1 & 2 & -3 & p \\ 2 & 6 & -11 & q \\ 1 & -2 & 7 & r \end{array}\right] = \left[\begin{array}{rrr|r} 1 & 2 & -3 & p \\ 0 & 2 & -5 & q-2p \\ 0 & 0 & -11 & r+2q-5p \end{array}\right] $$
Now we can see that the co-efficient matrix and augmented matrix, both has rank of 3 which equals the number of variables present i.e. $3$. So in this condition there must be unique solution to this system of simultaneous linear equations.
I am getting confused by the language of the question which says that there must be "at least one solution". So does this mean that this system of simultaneous linear equations should have more than 1 solutions? How can this be possible if the Gauss Elimination method is telling that there should be a unique solution?
Also what can we say about the relationship between $p$,$q$ and $r?$ As per my understanding, in order to keep the rank 3 because then only we will have a unique solution, so is the relationship between $p$,$q$ and $r$ will be $r+2q-5p = 0?$ My confusion here is that it is not necessary to have $r+2q-5p = 0$ because even if this is not equal to $0$ then also the rank will remain as $3$ as $-11$ will be there in the $3rd$ row.
Please clarify me on my doubts!
Your restriction is on the last row. If $r + 2q - 5p \neq -11z$ there is no solution.
EDIT: Upon double checking, I believe you made an error on your Gaussian Elimination. The last row should have $0 = r + 2q - 5p$. So that is your restriction.