Let $A=\begin{pmatrix} 1 &3 &1 &1 \\ 2 &2 &-1 &0 \\ 2 &4 &-2 &0 \\ 0 &-1 &2 &m \end{pmatrix}$
Find $m$ so that $r(A^{-1})=2$.
I have posted a similar problem before and they told me try to find $m$ such that $A$ is invertible. But i still feel it isn't enough. With the condition of $m$ such that $A$ is invertible ( $m\neq \frac{5}{3}$ ), how could we know that $r(A^{-1})=2$.
There is no way for the rank of $A^{-1}$ to be $2$. As you know for the existence of $A^{-1}$ we must have $A$ is non singular, in that case $r(A)=4$. Also we know if $A$ is invertible then so is $A^{-1}$, i.e., $A^{-1}$ will also become non-singular. Hence it's rank will also must be $4$.