Inverse of this $3\times 3$ matrix using the Cayley–Hamilton theorem Find the inverse of the matrix $$\begin{pmatrix} -1 & 2& 0
\\ 1& 1 &0 \\ 2 & -1& 2 \end{pmatrix}$$ using the Cayley–Hamilton theorem.
Thanks!
 A: Hint: The equation
$$
A^n + a_{n-1}A^{n-1} + \cdots + a_1 A + a_0 I  = 0
$$
can be rewritten as
$$
A(A^{n-1} + a_{n-1}A^{n-2} + \cdots + a_1I) = -a_0I
$$
A: The matrix $A$ is:
$A = \begin{bmatrix} -1 & 2 & 0 \\ 1 & 1 & 0 \\ 2 & -1 & 2 \end{bmatrix}, \tag{1}$
so the characteristic polynomial $p_A(\lambda)$ is
$p_A(\lambda) = \det(A - \lambda I) = \det \begin{bmatrix} -1 - \lambda & 2 & 0 \\ 1 & 1 - \lambda & 0 \\ 2 & -1 & 2 - \lambda \end{bmatrix}$
$= ( -1 - \lambda)(1 - \lambda) (2 - \lambda)  - 2(2 - \lambda) = (\lambda^2 - 1)(2 - \lambda) - 4 + 2\lambda$
$= -\lambda^3 + 2\lambda^2 + 3\lambda - 6, \tag{2}$
and by Cayley-Hamilton we have 
$0 = p_A(A) = -A^3 + 2A^2 + 3A - 6I; \tag{3}$
(3) may be written as
$A(-A^2 +2A + 3I) = 6I, \tag{4}$
or
$A(\dfrac{1}{6}(-A^2 +2A + 3I)) = I, \tag{5}$
which shows that
$A^{-1} = \dfrac{1}{6}(-A^2 +2A + 3I). \tag{6}$
I leave the explicit calculation of $A^{-1}$ from (6) to any interested readers; it is not difficult.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
