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

Question

Starting with the adjoint rule $$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)$$ So,

1. $(A^{-1})^{-1}=\dfrac{1}{\det(A^{-1})}\operatorname{adj}(A^{-1})$

2. $A=\dfrac{1}{\det(A^{-1})}\operatorname{adj}(A^{-1})$

3. $\det(A^{-1})A=\operatorname{adj}(A^{-1})$

4. $\dfrac{1}{\det(A)}A=\operatorname{adj}(A^{-1})$

Is this method simply done by considering the adjoint rule and then replacing all $A$ 's in the adjoint rule with $A^{-1}$? Technically that would be correct, no? Or is it doing something else?

Answer 1

@Bernard - "It also uses the multiplicative property of determinants. But basically , it uses this single rule."

Additionally,

@Michael Hoppe - "I In general the adjoint rule read for every matrix $det(A)⋅I=A⋅adj(A)$"