Identity matrix in adjoint proof

Question

Prove: If A is invertible, then adj(A) is invertible and $[adj(A)]^{-1}=\frac{1}{det(A)}A$

It starts using the adjoint rule

$$A^{-1}=\frac{1}{det(A)}adj(A)$$

$$AA^{-1}=\frac{1}{det(A)}A*adj(A)\longrightarrow I=\frac{1}{det(A)}A*adj(A)$$

and the proof goes on to say,

$$[adj(A)]^{-1}=\frac{1}{det(A)}A$$

But I'm not sure how to show (on the left side): $$\frac{I}{[adj(A)]}=[adj(A)]^{-1}$$ Why can $I$ be used interchangeably with $1$?

Answer 1

From $$I = \frac{1}{\det(A)} A \cdot adj(A)$$

We have shown that $adj(A)$ is invertible.

We then post-multiply the equation by $[adj(A)]^{-1}$ to obtain the result.

Note that $adj(A)$ is a matrix.

Answer 2

$I$ is the identity matrix and it follows that for any matrix $A$, $AI = IA=A$.