If is in the null space of A then we must have .

We must have 2a- b+ 3c+ 4d= 0, -a+ 6b+ 4c+ 9d= 0, 5a- 12b- c- 9d= 0. That is three equations in four unknowns. You should be able to solve for three of them, say, a, b, c in terms of d. In that case, replacing a, b, c, by those you can write the vector in terms of d only. Factor d out of that and you have d times a vector which is the basis vector.

It might happen that you the equations above are not independent. In that case you might be only able to solve for two of the unknowns in terms of the other two or only one in terms of the other three. In that case you can write the vector above as a sum of one variable times a vector plus the other times a vector. The basis vectors are those two vectors and similarly for three vectors if you can solve for only one of a, b, c, d in terms of the other three.

Multiply that matrix by each of the given vectors x and y. Yes, it really is that easy!2. What are the coordinate vectors of and relative to the basis ?