Getting characteristic polynomial from a small matrix Sorry I don't know how to format matrices, but if I have this matrix
$\pmatrix{1& 1& 0\\
    0& 0& 1\\
    1 &0& 1\\}$
How is the characteristic polynomial $λ^3 − 2λ^2 + λ − 1$? Is there some methodical approach to getting the characteristic polynomial from a matrix? 
EDIT:
$\text{Det}(A - \lambda I)$ means
$\pmatrix{1-\lambda& 1& 0\\
    0& -\lambda& 1\\
    1 &0& 1-\lambda\\}$
and so the determinant of this matrix is
$= (1-\lambda)((-\lambda)(1-\lambda) - (1)(0)) - (1)((0)(1-\lambda)-(1)(1)) + (0)((0)(0) - (-\lambda(1)))$
$= -\lambda^3+2 \lambda^2-\lambda+1$
Huh, seems to be similar, but the signs are different?
 A: In most elementary linear algebra courses you're looking for the characteristic polynomial as means to study the eigenvalues. Given an $n \times n$ matrix $A$,  $A$ has eigenvalue $\lambda \in \mathbb{R}$ if there exists $v \in \mathbb{R}^n$ such that $$Av = \lambda v$$ But then we have $(A - \lambda I)v = 0$. By definition, $A-\lambda I$ is then not invertible and hence it has $0$ determinant. So the eigenvalues are the solutions to the equation $\det(A-\lambda I) = 0$. The fact that the signs are different in your case doesn't matter: the solutions to any polynomial $p(x)=0$ are the same as the solutions to $-p(x)=0$.
A: As mentioned in the comments, you just find $\det(A-\lambda I)$ (or $\det(\lambda I-A)$ if you want the leading term positive).  Alternatively, if you find all of the (complex) eigenvalues $\lambda_1, \lambda_2, \lambda_3$, counted with multiplicity, then the characteristic polynomial will be $(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$.  In this particular case you'll want to do the first method because the roots of the characteristic polynomial of $A$ are pretty gnarly:

~ WolframAlpha
