Method to get Matrix inversion of special form $(sI-A)^{-1}$

Question

I am learning Control theory , and I often encounter the following question? $$(sI-A)^{-1}$$ for example when $$A=\begin{pmatrix} 2&3&1\\ 1&3&1\\ 1&2&2 \end{pmatrix}$$ I get $$(sI-A) = \begin{pmatrix} s-2&-3&-1\\ -1&s-3&-1\\ -1&-2&s-2\\ \end{pmatrix}$$ My idea is following $$B=\begin{pmatrix} s-2&-3&-1&1&0&0\\ -1&s-3&-1&0&1&0\\ -1&-2&s-2&0&0&1\\ \end{pmatrix}$$ Then do a series of transformation to get the inversion of the matrix , am I right? Is there a simpler method ?

Answer 1

The characteristic polynomial of $A$ is $$ p(\lambda)= \lambda^3-7\lambda^2+10\lambda+4. $$ Therefore $p(A)=0$. From this, \begin{align} -p(\lambda)I & =p(A)-p(\lambda)I \\ &=(A^3-\lambda^3I)-7(A^2-\lambda^2I)+10(A-\lambda I) \\ &= (A-\lambda I)(A^2+\lambda A+\lambda^2I)-7(A-\lambda I)(A+\lambda I)+10(A-\lambda I) \\ &= (A-\lambda I)(A^2+\lambda A+\lambda^2 I-7A-7\lambda I+10I) \\ &= (A-\lambda I)(\lambda^2 I+\lambda (A-7I)+A^2-7A+10I) \end{align} Hence, if $p(\lambda)\ne 0$, then $A-\lambda I$ is invertible, with inverse given by $$ (A-\lambda I)^{-1}=-\frac{1}{p(\lambda)}(\lambda ^2I+\lambda (A-7I)+A^2-7A+10I) $$