Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Digital systems, linear algebra, introduction to formal proof / 4 / b

Gaussian elimination givesElementary row operations preserve the row space, so $\{(1,0,1),(0,1,2)\}$ is a basis. The two rows are linearly independent because their first two coordinates form the standard basis of ${\mathbb{R}}^2$.

The pivot columns of the reduced matrix in part (i) are columns one and two. Select those columns of the original matrix, giving the column-space basisThey are independent, for example their first two entries have determinant $-2\ne 0$. The remaining original column is the first plus twice the second. Taking the reduced matrix's columns instead would generally give the wrong column space.

The reduced equations are $x+z=0$ and $y+2z=0$. Thus the null space is the lineThe nonzero spanning vector gives a one-element basis, so the nullity is one.