Solution

All that was required of the matrix A is that its nullspace be a particular vector space. That is, that its row space

be a particular row space. Thus we can change A by any row operations since this doesn't change the row space.

If we make the problem even more interesting by insisting that we want the vector to stay the same while getting infinitely many choices for , then we proceed as follows.

We take one of the rows of , say the first and add to it a multiple of a (row) vector which is simultaneously perpendicular to all three vectors . It is not hard to see that most choices of such a multiple will lead to the desired answer.

So this is how we do it! Let H be the matrix which has added as a third row.

Calculating the null space of this matrix finds all vectors perpendcular to all three of these. The matrix has

rank 3 so the nullspace has rank 1. We calculate that it has v = as a basis.

If we subtract a multiple of v from a row of A then the resulting matrix times any of the vectors in the previous problem will yield the same answer as when it is multiplied by A (Be sure to check this)

In fact if we subtract t*v from the first row of A we get

Since we want to be sure that the is the entire solution we need to be sure that the matrix is still rank 2 (otherwise the null space will acquire another dimension). If t = 1/3 this happens so we would need to avoid that value.