In part (a) we found that
![{\displaystyle A={\begin{bmatrix}12&4&1&0\\0&0&1&0\end{bmatrix}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3380188c6679f8af0cd995f7bad60849a1e5318e)
The matrix A has 2 pivot columns and so by rank-nullity theorem
![{\displaystyle \displaystyle {}\dim(N(A))=4-\dim(R(A))=4-2=2}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6429fd0292cbfce5b4ce930dac7d4122bda8775d)
and so there are two basis vectors for the nullspace. To find them we could row reduce A or notice that, at x = 0,
![{\displaystyle \displaystyle {}3a_{1}(0)^{2}+2a_{2}(0)+a_{3}=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9a5240be255d8c0d5a58780aac1b78eb7f44b167)
gives
. Using the value at x=2
![{\displaystyle \displaystyle {}12a_{1}+4a_{2}+a_{3}=12a_{1}+4a_{2}=0,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/38b481bcdb265c709ceed0cb351e94807dd65ac5)
gives
. Since neither equation depends on
then it is a free variable. Hence, the basis of N(A) is
Hence,