# Proof of the inverse of a matrix multiplication from the relation $\operatorname{inv}(A) =\operatorname{adj}(A)/\det(A)$

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)?

• Why would you want to prove such a simple one using the classic adjoint? Won't it be simpler to use uniqueness of inverses? – DonAntonio Feb 20 at 18:29
• (1) is a general statement valid in any (multiplicative) group. – Bernard Feb 20 at 18:57
• My question is really how to prove (3) – 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 $$\operatorname{adj}(AB) = \operatorname{adj}(B)\operatorname{adj}(A)$$.

• 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. – DonAntonio Feb 20 at 18:38
• 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. – Watercrystal Feb 20 at 18:43
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.
• 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... – DonAntonio Feb 20 at 21:33