0
$\begingroup$

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)?

$\endgroup$
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
1
$\begingroup$

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 $\operatorname{adj}(AB) = \operatorname{adj}(B)\operatorname{adj}(A)$.

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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

$$(AB)\cdot(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.