Hello everyone.
The matrix is
1 1 -1 2
2 2 -3 1
-1 -1 0 5
I calculated the null space of the matrix as
Yet the answer is supposedly
Why can't I express the first as part of the linear combination (i.e. as part of the span)?
You get exactly the same result. That matrix row reduces to
so that (x, y, z, u) will be in the null space if and only if x+ y- z+ 2u= 0, z+ u= 3, 6u= 0. Those last two equations give z= u= 0 and the first equation reduces to x+ y= 0 or x= -y. Every vector in the null space is (x, -x, 0, 0)= x(1, -1, 0, 0). The null space is one-dimensional and {(1, -1, 0, 0)} is a basis ((-1, 1, 0, 0) is a multiple of that so it would also be a basis vector.