For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $A\prec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.

If $$ B = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}^{\otimes10}, $$ then the matrix $$ A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}^{\otimes10} $$ satisfies $\mathrm{rk}(A)<\mathrm{rk}(B)$ and $I\prec A\prec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $\mathrm{rk}(C)<\mathrm{rk}(A)$ and $I\prec C\prec A$?