Consider the matrix
$A$ = $\begin{pmatrix}1&0&1&-1\\ 2&-1&0&-5\\ 0&1&2&3\end{pmatrix}$
(a) Determine the rank of A.
(b) Let r = rank(A). Find invertible matrices U and V such that
$U$$AV$ = $\begin{pmatrix}Ir&0\\ 0&0\end{pmatrix}$
So after row reducing the matrix, I figured out that the rank was $2$. But I am not sure how I will find the invertible matrices $U$ and $V$. I understand that it will require column operations but I am not sure how specifically. I do know that $Ir$ is the $2$x$2$ identity matrix but I am not sure how I would use it here.
Any assistance would be highly appreciated!
Here's a neat approach: take the "augmented matrix" $$ \left(\begin{array}{c|c}A & I\\ \hline I & 0\end{array}\right) = \left(\begin{array}{cccc|ccc}1&0&1&-1&1&0&0\\ 2&-1&0&-5&0&1&0\\ 0&1&2&3&0&0&1 \\ \hline 1 & 0 &0 &0 &0&0&0\\ 0&1 & 0 &0 &0 &0&0\\ 0&0&1 & 0 &0 &0 &0\\ 0&0&0&1 & 0 &0 &0 \end{array}\right) $$ Apply row/column operations until $A$ is put into the desired form. The resulting matrix will be $$ \left(\begin{array}{c|c}M & U\\ \hline V & 0\end{array}\right) $$ That is: the upper-right block will be your $U$, and the lower-left block will be your $V$. Note that your row-operations should only affect $U$, and your column-operations should only affect $V$.