Actually the characteristic polynomial is often defined as
$$
\chi_A=\det(I_nX-A)\in k[X]
$$
so as to be always monic (of degree $n$); see for instance in Wikipedia. This differs by a sign (and by calling the identity matrix by the more ususal name of $I_n$) from the definition you cited. The fact that the two contradicting definitions coexist shows that the matter of a factor $(-1)^n$ is not considered of great importance.
In my experience however, in most applications of the characteristic polynomial other than just for searching eigenvalues, the fact that it is monic is of importance. (One such application is to show that certain values are algebraic integers over the base ring, i.e., solutions of a monic polynomial equation.) For sure, in those applications monic-up-to-a-sign will be easily seen to do the job as well, but it is more convenient if the characteristic polynomial is just monic, period.
Also consider the statement "the coefficient of degree $n-i$ of $\chi_A$ is the $i$-th symmetric function of minus the eigenvalues of $A$ (taken with thir algebraic multiplicities)". With the definition you gave, you'd need to throw in another "minus".
Once upon a less enlightened time, when people were less knowledgeable in the intricacies of algorithmically computing eigenvalues, methods for generating the coefficients of a matrix's eigenpolynomial were quite widespread. One of the more prominent methods for computing the coefficients was a method ascribed to both the Frenchman Leverrier, and the Russian Faddeev (who was an (co-)author of one of the oldest references on the practice of numerical linear algebra).
The (Faddeev-)Leverrier method is a method that will require you to do a number of matrix multiplications to generate the coefficients of the characteristic polynomial. Letting the $n\times n$ matrix $\mathbf A$ have the monic characteristic polynomial $(-1)^n \det(\mathbf A-\lambda\mathbf I)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0$, the algorithm proceeds like so:
$\mathbf C=\mathbf A;$
$\text{for }k=1,\dots,n$
$\text{if }k>1$
$\qquad \mathbf C=\mathbf A\cdot(\mathbf C+c_{n-k+1}\mathbf I);$
$c_{n-k}=-\dfrac{\mathrm{tr}(\mathbf C)}{k};$
$\text{end for}$
If your computing environment can multiply matrices, or take their trace (sum of the diagonal elements, $\mathrm{tr}(\cdot)$), then you can easily program (Faddeev-)Leverrier. The method works nicely in exact arithmetic, or in hand calculation (assuming you have the stamina to repeatedly multiply matrices), but is piss-poor in inexact arithmetic, as the method tends to greatly magnify rounding errors in the matrix, ever yielding coefficients that become increasingly inaccurate as the iteration proceeds. But, for the simple $3\times 3$ case envisioned by the OP, this should work nicely.
People interested in this old, retired method might want to see this paper.
Best Answer
To flesh out the comment a little bit more: by definition, the characteristic polynomial of a matrix $A$ equals $\det(xI-A)$ which is seen to be a monic polynomial, so the more interesting question is whether every monic polynomial $P(X)\in R[X]$ is a characteristic polynoimal of some matrix with coefficients in $R$. And the answer to this question is yes, with an explicit construction given by the companion matrix.
This means that we associate to the polynomial $P(X)=X^n+c_{n-1}X^{n-1}+\cdots+c_0$ the matrix $$A(c_0,c_1,\ldots,c_{n-1}):= \begin{pmatrix} 0&\ldots &0&-c_0\\ 1&\ldots &0&-c_1\\ \vdots&\ddots &\vdots&\vdots\\ 0&\ldots &1&-c_{n-1}\\ \end{pmatrix}=\begin{pmatrix}0&-c_0\\I_{n-1}&-c_{1:n-1}\end{pmatrix} $$ with the coefficients $c_0,\ldots,c_{n-1}$ elements of $R$, where the rightmost expression is using a block matrix notation: $I_{n-1}$ is the $(n-1)\times (n-1)$ identity matrix, and $c_{1:n-1}$ is the column vector in $R^{n-1}$ with entries $(c_1,\ldots,c_{n-1})$.
Now we can compute the characteristic polynomial of $A(c_0,c_1,\ldots,c_{n-1})$ using cofactor expansion along the top row, as follows (using $\operatorname{charp}$ to denote the characteristic polynomial of a matrix):
$$\begin{align*} \operatorname{charp} A(c_0,c_1,\ldots,c_{n-1})&=\det\Bigl[X\ I_n-\begin{pmatrix}0&-c_0\\I_{n-1}&-c_{1:n-1}\end{pmatrix}\Bigr]\\ &=X\det\Bigl[X\ I_{n-1}-\begin{pmatrix}0&-c_1\\I_{n-2}&-c_{2:n-1}\end{pmatrix}\Bigr]+c_0\det I_{n-1}\\ &=X \operatorname{charp} A(c_1,\ldots,c_{n-1})+c_0. \end{align*} $$ Thus, it follows by induction on $n$ that $\operatorname{charp} A(c_0,c_1,\ldots,c_{n-1})=P(X)$.