# If $\det(A) =\det(B)$ $\det(\operatorname{adj}(A)) = \det(\operatorname{adj}(B))$

Given two matrices $$A, B$$ of size $$n \times n$$ where $$n \geqslant 2$$

If $$\det(A) =\det(B)$$, can we infer that $$\det(\operatorname{adj}(A)) = \det(\operatorname{adj}(B))$$?

If so, how can we prove this?

My proof was intuitive, but I feel like it would not suffice.

$$\det(\operatorname{adj}(A)) = \det(A)^{n-1}$$ therefore... it is the same as $$\det(B)^{n-1}$$ and case closed.

But I think the whole purpose was to prove that $$\det(\operatorname{adj}(A)) = \det(A)^{n-1}$$ which I have no idea how to prove.

$$A.adj(A)=det(A)I_n$$

$$det(A.adj(A))=det(det(A)I_n)$$

$$det(A).det(adj(A))=det(A)^ndet(I_n)$$

$$det(A).det(adj(A))=det(A)^n.1$$

$$\Rightarrow det(adj(A))=\frac {det(A)^n} {det(A)}$$

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

• Oh wow, didn't even think of that. Thank you! – Alex Osheter Mar 12 '20 at 11:51
• @Alex You are welcome – Mathsmerizing Mar 12 '20 at 11:51
• You have to be careful to avoid dividing by $0$ in the case $\det(A) = 0$, though. Use the fact that $\det(A)$ and $\det(\text{adj}(A))$ are polynomials in the entries of $A$, and take a limit. – Robert Israel Mar 12 '20 at 12:29