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The classical adjoint $\operatorname{adj}(A)$ of a square matrix $A$ has its $(i,j)$-th entry equal to the $(j,i)$-th cofactor (signed minor) of $A$. If $\det(A)\neq0$ we can define the inverse $A^*$ of $A$ as $\operatorname{adj}(A)/\det(A)$ and use $\operatorname{adj}(A)=\det(A)A^*$. It is known that, for arbitrary $n\times n$ matrices $A$ and $B$, $\operatorname{adj}(AB)=\operatorname{adj}(B)\operatorname{adj}(A)$. The property is either not stated in popular Linear Algebra books or, if stated, it is only proven for invertible matrices by using the properties: \begin{align*} \operatorname{adj}(AB) &= \det(AB)(AB)^*\\ &= (\det A)(\det B)B^*A^*\\ &= (\det(B)B^*)(\det(A)A^*)\\ &= \operatorname{adj}(B)\operatorname{adj}(A). \end{align*} MY POINT is that no new proof is required when $A$ and $B$ are not necessarily invertible. Just look at the equality as at a collection of n.n polynomial type identities, each in terms of n.n entries of and $A$ and n.n entries of $B$. It is enough to consider generic matrices $A$ and $B$, having independent variables as entries. In such a case $A$ and $B$ have nonzero determinants and so they are invertible. Thus, the proof mentioned above applies and the identity must be true for each pair of $n\times n$ matrices $A$ and $B$. My method would also apply to a few other adjoint identities, which are true in general but only proven for invertible matrices. A few of my colleagues (teachers) could not easily agree with my approach although one finally did. What is your opinion ?

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This is a known argument (good job finding it!): see the two "Universal identities" papers on math.uconn.edu/~kconrad/blurbs . –  darij grinberg May 13 '13 at 23:47
To reiterate your argument, we can identify $X=Mat_n$ with the affine space $\mathbb{A}^{n^2}$, and the ajdoint map is a morphism from $X$ to $X$ that sends the open and Zariski dense subset $U=GL_n$ to itself. We have two morphism $(A,B)\mapsto (AB)^*$ and $(A,B)\mapsto B^*A^*$ which agree on $GL_n\times GL_n$, and by continuity they agree on the whole $X\times X$ since $U\times U$ is dense in $X\times X$. –  Jiangwei Xue May 16 '13 at 8:07
Your notation is not quite clear to me but, anyhow, how can you talk about continuity and dense sets in the of an arbitrary integral domain ( a commutative ring with no zero divisors) ? –  Christopher Sokolnicki May 27 '13 at 21:44
@ChristopherSokolnicki The Zariski topology can be defined on $F^k$ where $F$ is an arbitrary field. Jiangwei Xue's method applies to fields, but you can always take the quotient field $Q(R)$ of your domain $R$ and observe that $\mathrm{adj}(A)$ has coefficients in $R$ if $A$ has. –  egreg May 27 '13 at 22:15
Just: why do you denote the inverse $A^*$ and not $A^{-1}$? –  1015 May 27 '13 at 22:58
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Let $R$ be a (commutative) domain and consider its quotient field $F$; build the field $F(\mathcal{X})$ where $\mathcal{X}$ is a set of the $2n^2$ indeterminates $X_{11},\dots,X_{1n},\dots,X_{n1},\dots,X_{nn}$ and $Y_{11},\dots,Y_{1n},\dots,Y_{n1},\dots,Y_{nn}$. Then the matrices $\mathbb{A}=[X_{ij}]$ and $\mathbb{B}=[Y_{ij}]$ are obviously invertible in $F(\mathcal{X})$ because their columns are certainly linearly independent.

Notice now that $\def\adj{\mathrm{adj}}\adj(\mathbb{A})$, $\adj(\mathbb{B})$ and $\adj(\mathbb{AB})$ have their coefficients in $R[\mathcal{X}]$ (the polynomial ring). Then


because the matrices are invertible.

Now, if you have two matrices $A[a_{ij}]$ and $B=[b_{ij}]$ with coefficients in $R$, just specialize the above relation with the evaluation $X_{ij}\mapsto a_{ij}$ and $Y_{ij}\mapsto b_{ij}$.

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