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

From part (a), we found that

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

And from part (a), we saw that only the column 1 and column 3 are pivot columns. Thus column 1 and 3 of matrix A are linearly independent and span R(A). So a basis of R(A) is given by

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