Let

$$A = \begin{bmatrix} x_{11} A_{11} & x_{12} A_{12} & x_{13} A_{13} & \cdots & x_{1d} A_{1d}\\ x_{21} A_{21} & x_{22} A_{22} & x_{23} A_{23} & \cdots & x_{2d} A_{2d}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{d1} A_{d1} & x_{d2} A_{d2} & x_{d3} A_{d3} & \cdots & x_{dd} A_{dd}\\ \end{bmatrix}$$

where $A_{ij}$ is an $n \times n $ invertible matrix and $x_{ij} \neq 0$ is a scalar. All the entries in matrices and scalars are from a finite field $\mathbb{F}_q$. Let $X = (x_{ij})$. It can easily be shown that there exists a matrix $X$ such that the matrix $A$ is invertible if $q > 2$. But is there any nice sufficient condition I can put on matrix $X$ or an explicit construction of $X$ to make $A$ invertible other than saying the $\det(A)$ polynomial in $x_{ij}$ must be non-zero?

"other than saying the $\det(A)$ polynomial in $x_{ij}$ must be non-zero"- careful here, because $A$ is a matrix over $M_n (\Bbb F_q)$ which is not commutative. You need first a notion of determinant for matrices with entries in a non-commutative ring; there are several approaches to define one, one of them being the concept of "quasideterminant". $\endgroup$ – Alex M. Aug 10 '17 at 17:58