$$A=\begin{bmatrix}
3&1 &1 \\
1&3& 1\\
1&1&3
$
\end{bmatrix}$$
First part of problem was to find matrix $S$ such that $SDS^T=A$. Very easy find characteristic polynomial $k_A(\lambda)=-\lambda^3+9\lambda^2-24\lambda+20$.
$k_A(\lambda)=0$ iff $\lambda=5, \lambda=2$ and one more $\lambda=2$, and than find eig. vectors...
Second part of problem is to find inverse of A-3I, but you must use part a) in your solution.
My solution:
First idea was to use HC theorem:
$-A^3+9A^2-24A+20I=0$ and rewrite this polinomial as a sums of $(A-3I)$ (Taylor series at point 3)
$$2I+3(A-3I)-(A-3I)^3=0$$
$T=A-3I$
$$T^3-3T-2I=0$$
$$T^3-3T=2I$$
$$T(T^2-3I)=2I$$
$$T(\frac{T^2-3I}{2})=I$$
By definition of inverse all inside brackets is inverse of $T=A-3I$.
My question is there any more "linear algebra" approach in this problem.