Note in the following that the scalar multiple of a matrix is equivalent to multiplying each of the rows by that scalar, so we actually apply Theorem DRCM multiple times below (and are passing up an opportunity to do a proof by induction in the process, which maybe you'd like to do yourself?).
\begin{align*}
\charpoly{A}{x}
&=\detname{A-xI_n}&&\knowl{./knowls/definition.CP.knowl}{\text{Definition CP}}\\
&=\detname{(-1)(xI_n-A)}&&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\
&=(-1)^{n}\detname{xI_n-A}&&\knowl{./knowls/theorem.DRCM.knowl}{\text{Theorem DRCM}}\\
&=(-1)^{n}r_A(x)
\end{align*}
Since the polynomials are scalar multiples of each other, their roots will be identical, so either polynomial could be used in Theorem EMRCP.
Computing by hand, our definition of the characteristic polynomial is easier to use, as you only need to subtract $x$ down the diagonal of the matrix before computing the determinant. However, the price to be paid is that for odd values of $n$, the coefficient of $x^{n}$ is $-1$, while $r_A(x)$ always has the coefficient $1$ for $x^{n}$ (we say $r_A(x)$ is “monic.”)