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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?

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  • $\begingroup$ If the Aij are all the identity matrix, (and the xij are also the same), I do not see how A can have determinant different from 0. Indeed, even if the identity matrix is only for I=1,2 and all j (or the transpose) and similarly for the xij, I see A as being singular. Gerhard "And Q Doesn't Matter Here" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 10 '17 at 16:59
  • $\begingroup$ @Balajisb: "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
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    $\begingroup$ @AlexM. Presumably what the OP means is $\det(A)$ where $A$ is considered as a $dn \times dn$ matrix over $\mathbb F_q$. $\endgroup$ – Robert Israel Aug 10 '17 at 20:21
  • $\begingroup$ @AlexM as Robert pointed out i do mean $det(A)$ when A is considered as $dn \times dn$ matrix. $\endgroup$ – Balaji sb Aug 11 '17 at 6:30
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The only easy case I can think of is where $\mathrm A_{ij} = \mathrm A_0$, where $\mathrm A_0$ is an invertible $n \times n$ matrix, for all $i,j \in [d]$. In this special case,

$$\mathrm A = \begin{bmatrix} x_{11} \mathrm A_{0} & x_{12} \mathrm A_{0} & \cdots & x_{1d} \mathrm A_{0}\\ x_{21} \mathrm A_{0} & x_{22} \mathrm A_{0} & \cdots & x_{2d} \mathrm A_{0}\\ \vdots & \vdots & \ddots & \vdots \\ x_{d1} \mathrm A_{0} & x_{d2} \mathrm A_{0} & \cdots & x_{dd} \mathrm A_{0}\\ \end{bmatrix} = \mathrm X \otimes \mathrm A_0$$

where $\otimes$ denotes the Kronecker product. Hence,

$$\det (\mathrm A) := \left( \det (\mathrm X) \right)^n \cdot \left( \det (\mathrm A_0) \right)^d$$

and, thus, $\mathrm A$ is invertible if and only if $\mathrm X$ is invertible. If that is indeed the case, the inverse is

$$\rm A^{-1} = X^{-1} \otimes A_0^{-1}$$

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    $\begingroup$ Even if the matrix are non-square, I think you can take an SVD $X=U_1S_1V_1^*$ and $A_0=U_2S_2V_2^*$ and combine them as $X\otimes A_0 = (U_1 \otimes U_2) (S_1\otimes S_2) (V_1^*\otimes V_2^*)$ and reduce to the diagonal case. $\endgroup$ – Federico Poloni Sep 11 '17 at 6:58

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