For A= $\begin{bmatrix} 4&3&1 \\ 2&1&-2 \\ 1&2&1\\ \end{bmatrix}$

Hence find $A^{-1}$

$\begin{align} A=\begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix}\ \lvert A \lvert = 11 \ \text{Characterstic equation is given by}\ \lvert A-\lambda I \lvert = 0\ \begin{bmatrix} &\ 4-\lambda &3&1 \ 2&1-\lambda &-2 \ 1&2&1-\lambda \ \end{bmatrix} = 0\ \end{align}$ $\begin{align} \lambda ^3 - 6\lambda ^2 +(4+1+4-6-1+4)\lambda - 11 & =0 \ \lambda ^3 - 6\lambda ^2 + 6\lambda ^3 - 6\lambda ^2 -11 &=0 \ \text{by cayley Hamilton theorem, this equation is satisfied by A} \ A^3-6A^2+6A-11I & = 0 - - - -1 \ \end{align}$ $\begin{align} A&=\begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix} - - - - 2\ A^2= A.A& = \begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix} \begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix} \ A^2 &=\begin{bmatrix} &\ 23&17&-1 \ 8&3&-2 \ 9&7&-2\ \end{bmatrix}\ A^3= A^2.A& = \begin{bmatrix} &\ 23&17&-1 \ 8&3&-2 \ 9&7&-2\ \end{bmatrix} \begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix}\ A^3&= \begin{bmatrix} \ 125&84&-12 \ 36&23&0 \ 48&30&-7\ \end{bmatrix} - - - - 4\ I &= \begin{bmatrix} \ 1&0&0 \ 0&1&0 \ 0&0&1\ \end{bmatrix} - - - - 5\ \text{put equation 2,3,4,5 in euation 1}\ \end{align}$ $\begin{align} \begin{bmatrix} \ 125&84&-12 \ 36&23&0 \ 48&30&-7\ \end{bmatrix} - 6 \begin{bmatrix} &\ 23&17&-1 \ 8&3&-2 \ 9&7&-2\ \end{bmatrix} + 6 \begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix} -11 \begin{bmatrix} \ 1&0&0 \ 0&1&0 \ 0&0&1\ \end{bmatrix} = 0 \ \text{Hence cayley Hamilton theorem is verified.}\ \end{align}$ $ \begin{align} \text{To find $A^{-1}$, multiply equation 1 by $A^{-1} , we get $}\ A^3-6A^2+6I-11A^{-1} & = 0\ 11A^{-1} & = A^3-6A^2+6I - - - - 6\ \text{put equation 2,3,4,5 in equation 6}\ \end{align}$ $\begin{align} 11A^{-1} & = \begin{bmatrix} &\ 23&17&-1 \ 8&3&-2 \ 9&7&-2\ \end{bmatrix} - 6 \begin{bmatrix} &\ 4&3&1 \ 2&1&-2 \ 1&2&1\ \end{bmatrix} +6 \begin{bmatrix} \ 1&0&0 \ 0&1&0 \ 0&0&1\ \end{bmatrix}\ 11A^{-1} & = \begin{bmatrix} &\ 5&-1&-7 \ -4&3&10\ 3&-5&-2\ \end{bmatrix} \ A^{-1} & = \frac{1}{11} \begin{bmatrix} &\ 5&-1&-7 \ -4&3&10\ 3&-5&-2\ \end{bmatrix} \end{align}$