Yes, that is correct. Form the "augmented" matrix consisting of Q with b1, b2, b3, b4, b5 forming the 5th column. Row reduce. You will always be able to reduce the first 4 number on the last row to 0. (And, if the equations are not independent, perhaps on higher rows.) The last column will be, of course, combinations of the "b"s. The equation will have a solution if and only if the numbers in the last column for "all 0 rows" are also 0. Those will be be conditions on the "b"s.