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By the formulas in the article, we can use the formula \begin{align} A^{-1} = \frac{1}{\det(A)}\operatorname{adj}(A). \end{align} where $\operatorname{adj}(A)$ is a matrix with $(i,j)$-entry: $(-1)^{i+j}{M}_{ji}$, $M_{ji}$ is the determinant of the matrix obtained from $A$ by deleting the $j$-th row and $i$-th column.

Does this formula work for matrices with non-commutative entries? Let $A=(a_{ij})_{n \times n}$, $a_{ij}'s$ do not commute. I use the following definition of determinant for a matrix with non-commutative entries. $$ \det(A)=\sum_{\sigma \in S_n} (-1)^{\tau(\sigma)} a_{1,\sigma(1)} a_{2,\sigma(2)} \cdots a_{n,\sigma(n)}. $$ Do we still have \begin{align} A^{-1} = \det(A)^{-1} \operatorname{adj}(A)? \end{align} Thank you very much.

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  • $\begingroup$ I am rather sure you can find a $2\times 2$ counterexample (if not for matricial entries, at least in a free ring). A lot of things should go wrong, without commutativity. $\endgroup$ – user228113 May 3 '17 at 14:16
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Let's inspect the $2\times 2$ case $$(\det A)^{-1}(\operatorname{adj}A)A=\\=\begin{bmatrix}(a_{11}a_{22}-a_{12}a_{21})^{-1}a_{22}&-(a_{11}a_{22}-a_{12}a_{21})^{-1}a_{12}\\-(a_{11}a_{22}-a_{12}a_{21})^{-1}a_{21}&(a_{11}a_{22}-a_{12}a_{21})^{-1}a_{11}\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}=\\=\begin{bmatrix}(a_{11}a_{22}-a_{12}a_{21})^{-1}(a_{22}a_{11}-a_{12}a_{21})&(a_{11}a_{22}-a_{12}a_{21})^{-1}(a_{22}a_{12}-a_{12}a_{22})\\(a_{11}a_{22}-a_{12}a_{21})^{-1}(a_{21}a_{11}-a_{11}a_{21})&(a_{11}a_{22}-a_{12}a_{21})^{-1}(a_{11}a_{22}-a_{21}a_{12})\end{bmatrix}$$

Now, if you equate with $1$ and multiply both sides by $\det A$ on the left, you obtain the system $$\begin{cases}a_{22}a_{11}-\color{red}{a_{12}a_{21}}=a_{11}a_{22}-\color{red}{a_{12}a_{21}}\\ a_{22}a_{12}=a_{12}a_{22}\\a_{21}a_{11}=a_{11}a_{21}\\ \color{red}{a_{11}a_{22}}-a_{21}a_{12}=\color{red}{a_{11}a_{22}}-a_{12}a_{21}\\ a_{11}a_{22}-a_{12}a_{21}\text{ is invertible}\end{cases}\iff \begin{cases}a_{22}a_{11}=a_{11}a_{22}\\ a_{22}a_{12}=a_{12}a_{22}\\a_{21}a_{11}=a_{11}a_{21}\\a_{21}a_{12}=a_{12}a_{21}\\ a_{11}a_{22}-a_{12}a_{21}\text{ is invertible}\end{cases}$$

So you can see that, for that matrix to be even just a left-inverse, a lot of entries of $A$ must commute between each other. For it to be a right inverse it gets worse, because in $A((\det A)^{-1}\operatorname{adj} A)$ you end up with sums of stuff in the form $a_{ij}(\det A)^{-1} a_{kh}$.

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