$B=\operatorname{adj}(A)$, then how to show $(-t)^n \phi_A\left( \frac{\det(A)}{t} \right) = \det(A) \phi_B(t)$

Question

Let $A$ be an invertible $n\times n$ matrix and let $B= \operatorname{adj}(A)$ be the classical adjoint of $A$.

I want to show that the characteristic polynomials $\phi_A(t)$ and $\phi_B(t)$ of $A$ and $B$, respectively satisfy

\begin{align}(-t)^n \phi_A\left( \frac{\det(A)}{t} \right) = \det(A) \phi_B(t). \end{align}

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I know $A \operatorname{adj}(A) = \det(A) I_n$, and $\det(c A) = c^n \det(A)$, but having problem dealing with characteristic polynomials

Answer 1

$$A(B-tI) = \det(A)I-tA = -t\bigg(A - \frac{\det(A)}tI\bigg) \stackrel{\textrm{taking det}}\implies \det(A)\phi_B(t) = (-t)^n \phi_A \bigg( \frac{\det(A)}t \bigg).$$