Determinant of the inverse matrix

Question

I'm seeking for a proof of the following:

Let $A$ be an invertible matrix. Then the determinant of $A^{-1}$ equals: $$\left|A^{-1}\right|=|A|^{-1} $$

I don't know where to begin the proof. Any suggestions?

Answer 1

Hint: You know that $\det(AB)=\det(A)\det(B)$ for all $n\times n$ matrices $A,B$. So make a judicious choice for $B$...

Answer 2

Exploiting the commutativity of the determinant operation with multiplication is probably the easiest way. That said, here is another approach.

The determinant of a square matrix is equal to the product of its eigenvalues.

Now note that for an invertible matrix $\mathbf A$, $\lambda\in\mathbb R$ is an eigenvalue of $\mathbf A$ is and only if $1/\lambda$ is an eigenvalue of $\mathbf A^{-1}$. To see this, let $\lambda\in\mathbb R$ be an eigenvalue of $\mathbf A$ and $\mathbf x$ a corresponding eigenvector. Then, \begin{align*} \mathbf A\mathbf x=&\,\lambda\mathbf x\\ \Longrightarrow\qquad{\phantom{\lambda^{-1}}}\mathbf x=&\,\lambda\mathbf A^{-1}\mathbf x\\ \Longrightarrow\qquad\mathbf \lambda^{-1}\mathbf x=&\,\mathbf A^{-1}\mathbf x. \end{align*} That is, $\lambda^{-1}$ is an eigenvalue of $\mathbf A^{-1}$ corresponding to the same eigenvector $\mathbf x$. The other direction is analogous.

Hence, the determinant of $\mathbf A^{-1}$ is equal to the product of the eigenvalues of $\mathbf A^{-1}$, which is the product of the reciprocals of the eigenvalues of $\mathbf A$, which is just the reciprocal of the determinant of $\mathbf A$.

Answer 3

By interpreting the determinant as the (signed) ratio between the hypervolume of $f_A(\Gamma)$ and the hypervolume of $\Gamma$, where $\Gamma$ is a symplex associated with the canonical base and $f_A$ is the linear map associated with $A$, the claim is trivial, since: $$ \left(f_A\right)^{-1} = f_{A^{-1}}. $$ This is just the classic "measure-theoretic" proof of Binet's theorem.