I am trying to prove that (1) $(\mathbf{AB})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$ using the relation (2)$$ \mathbf{A}^{-1}=\text{adj}\mathbf{A}/\det \mathbf{A}$$ where $\operatorname{adj}\mathbf{A}$ is the adjugate of $\mathbf{A}$, and $\det\mathbf{A}\neq 0$

For the special case where both matrices are $n$ x $n$, I came down to try to prove that (3) $$\text{det}(\mathbf{AB})(i/j)=\sum_{k=1}^n\text{det}\mathbf{A}(i/k)\det\mathbf{B}(k/j)$$ where $\det\mathbf{A}(i/j)$ is the determinant of $\mathbf{A}$ minus row $i$ and column $j$.

I have checked the validity of this relation for a $2×2$ matrix, but cannot prove it in general. On the other hand, is there a more straightforward way of proving (1) using (2) without going through (3)?

  • $\begingroup$ Why would you want to prove such a simple one using the classic adjoint? Won't it be simpler to use uniqueness of inverses? $\endgroup$ – DonAntonio Feb 20 at 18:29
  • $\begingroup$ (1) is a general statement valid in any (multiplicative) group. $\endgroup$ – Bernard Feb 20 at 18:57
  • $\begingroup$ My question is really how to prove (3) $\endgroup$ – Al-C Feb 20 at 21:35

There is an easy way: Suppose that $A, B$ are $n \times n$ matrices such that $AB$ is invertible. This means that $\det(AB) \neq 0$ and hence, by the formula you gave we have $$ \begin{align} (AB)^{-1} &= \frac{\operatorname{adj}(AB)}{\det(AB)} \\ &= \frac{\operatorname{adj}(B) \cdot \operatorname{adj}(A)}{\det(B) \cdot \det(A)} \\ &= \frac{\operatorname{adj}(B)}{\det(B)} \cdot \frac{\operatorname{adj}(A)}{\det(A)} \\ &= B^{-1} A^{-1} \end{align} $$ by using $\operatorname{adj}(AB) = \operatorname{adj}(B)\operatorname{adj}(A)$.

  • $\begingroup$ One would have to explain how did you go from adj$(AB)\;$ in the first line to adj$(B)\,$adj$(A)$ in the second line. $\endgroup$ – DonAntonio Feb 20 at 18:38
  • $\begingroup$ It's a known property I specifically mentioned and I guessed that OP would know it -- if not, there is an elementary proof for this case on Wikipedia. $\endgroup$ – Watercrystal Feb 20 at 18:43
  • $\begingroup$ How do you get adj(AB)= adj(B)adj(A) $\endgroup$ – Al-C Feb 20 at 20:47
  • $\begingroup$ That is explained in the comment right above yours. $\endgroup$ – Watercrystal Feb 20 at 20:59
  • $\begingroup$ The only elementary proof on Wikipedia I see uses the fact that inv(AB)=inv(B)inv(A), which is what we are trying to prove here. The other proof makes a reference to the Binet-Cauchy relation. $\endgroup$ – Al-C Feb 20 at 21:31

For any invertible matrix $\;X$, its inverse is denoted by $\;X^{-1}$. Thus, the inverse of the product $\;AB\;$ of square matrices $\;A,\,B\;$ is $\;(AB)^{-1}$. Yet


using associativity, and thus also $\;B^{-1}A^{-1}\;$ is the inverse of $\;AB$. By uniqueness of the inverse, $\;(AB)^{-1}=B^{-1}A^{-1}\;$ and we're done.

  • $\begingroup$ I am familiar with this proof, I was looking for a proof using (2). Also, if either det(A) or det(B)=0, then det(AB)=0 and AB has no inverse. How does the above proof covers this case? $\endgroup$ – Al-C Feb 20 at 21:01
  • $\begingroup$ The basic assumption, of course, is that both $\;A,\,B\;$ are invertible. It is exactly the very same assumption done with the other method, as you must divide by the determinant... $\endgroup$ – DonAntonio Feb 20 at 21:33

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