Science:Math Exam Resources/Courses/MATH307/December 2006/Question 02 (d)/Solution 1

From part (a), we found that

${\text{rref}}(A)={\begin{bmatrix}1&2&0&3&4\\0&0&1&-2&-1\\0&0&0&0&0\end{bmatrix}}$

We know that the basis of $R(A^{T})$ i.e columns of $A^{T}$ corresponding to the pivot columns of $rref(A^{T})$, is the same as the pivot rows of $rref(A)$. The pivot rows of $rref(A)$ are row 1 and 2.

Therefore a basis of R(A^T) is given by

$\left\{{\begin{bmatrix}1\\2\\0\\3\\4\end{bmatrix}},{\begin{bmatrix}0\\0\\1\\-2\\1\end{bmatrix}}\right\}$