Use Cayley Hamilton theorem to derive the following $(sI−A)^{-1}$ where $s \in \mathbb C$ .

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$(sI−A)^{-1}$ where $s\in\mathbb C$.

My attempt : I have seen this method of matrix equation using the Cayley Hamilton method .

$f(A)= (sI−A)^{-1} = h_0(I) + h_1(A)+ h_2(A)^2$;

$f(p) = h(p) = h_0 + h_1(p)+ h_2(p)^2$, where $p$ is the eigenvalues of $A$.

Given $A=\pmatrix{0&1&0\\0&0&1\\-6&-11&-6}$ the eigenvalues are $-1,-2,-3$.

$(s - (-1))^{-1} = h_0 + h_1(-1) + h_2(-1)^2$.

$1/(s+1) = h_0 - h_1 + h_2$.

Similarly for the other two eigenvalues, and solve for $h_0,h_1,h_2$.

Is this correct? Why is $s$ given as a complex number ?