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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.

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

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

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

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

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

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

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  • $\begingroup$ Oh wow, didn't even think of that. Thank you! $\endgroup$ – Alex Osheter Mar 12 '20 at 11:51
  • $\begingroup$ @Alex You are welcome $\endgroup$ – Mathsmerizing Mar 12 '20 at 11:51
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Mar 12 '20 at 12:29

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