0
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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

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

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

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

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

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

$\endgroup$
3
  • 1
    $\begingroup$ Oh wow, didn't even think of that. Thank you! $\endgroup$ Mar 12, 2020 at 11:51
  • $\begingroup$ @Alex You are welcome $\endgroup$ Mar 12, 2020 at 11:51
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 12, 2020 at 12:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .