Science:Math Exam Resources/Courses/MATH307/December 2012/Question 01 (b)/Solution 1

In part (a) we found that

$A={\begin{bmatrix}12&4&1&0\\0&0&1&0\end{bmatrix}}$

The matrix A has 2 pivot columns and so by rank-nullity theorem

$\displaystyle {}\dim(N(A))=4-\dim(R(A))=4-2=2$

and so there are two basis vectors for the nullspace. To find them we could row reduce A or notice that, at x = 0,

$\displaystyle {}3a_{1}(0)^{2}+2a_{2}(0)+a_{3}=0$

gives $a_{3}=0$. Using the value at x=2

$\displaystyle {}12a_{1}+4a_{2}+a_{3}=12a_{1}+4a_{2}=0,$

gives $a_{2}=-3a_{1}$. Since neither equation depends on $a_{4}$ then it is a free variable. Hence, the basis of N(A) is

${\vec {a}}={\begin{bmatrix}a_{1}\\a_{2}\\a_{3}\\a_{4}\end{bmatrix}}={\begin{bmatrix}a_{1}\\-3a_{1}\\0\\a_{4}\end{bmatrix}}={\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}}a_{1}+{\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}a_{4}$

Hence, ${\vec {a_{1}}}={\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}},{\vec {a_{2}}}={\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}$