My textbook gives the following problem:
Consider the block matrix $$A = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ 0 & 0 & A_{23} \end{bmatrix},$$ where $A_{11}$ is an invertible matrix. Determine the rank of $A$ in terms of the ranks of the blocks $A_{11}, A_{12}, A_{13}$, and $A_{23}$.
I've checked the answer in a solution manual and it claims $$\text{rref}(A) = \begin{bmatrix} I_p & A_{12} & * \\ 0 & 0 & \text{rref}(A_{23}) \end{bmatrix},$$ and $\text{rank}(A) = p + \text{rank}(A_{23}) = \text{rank}(A_{11}) + \text{rank}(A_{23})$. ($A_{11}$ is assumed to be $p \times p$.) I think I agree with the answer for $\text{rank}(A)$, but not $\text{rref}(A)$. When the row operations used to convert $A_{11}$ to $I_p$ are carried out on the first row of blocks, shouldn't $A_{12}$ and $A_{13}$ become $A_{11}^{-1} A_{12}$ and $A_{11}^{-1} A_{13}?$ My understanding is that further row operations would be required (after reducing $A_{11}$ to $I_p$) to handle the columns that contain $A_{13}$ and $A_{23}$, so that leading 1's in those columns are the only nonzero entries in those columns. So am I correct in thinking that $$\text{rref}(A) = \begin{bmatrix} I_p & A_{11}^{-1} A_{12} & * \\ 0 & 0 & \text{rref}(A_{23}) \end{bmatrix} \quad ? $$