1. Find Inverse of matrix using Cayley Hamilton method
`A=[[3,1,1],[-1,2,1],[1,1,1]]`

Solution:
To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial p(t) of the matrix A.
`|A-tI|`

=	`(3-t)`   `1`   `1`   `-1`   `(2-t)`   `1`   `1`   `1`   `(1-t)`

`=(3-t)((2-t) × (1-t) - 1 × 1)-1((-1) × (1-t) - 1 × 1)+1((-1) × 1 - (2-t) × 1)`

`=(3-t)((2-3t+t^2)-1)-1((-1+t)-1)+1((-1)-(2-t))`

`=(3-t)(1-3t+t^2)-1(-2+t)+1(-3+t)`

`= (3-10t+6t^2-t^3)-(-2+t)+(-3+t)`

`=-t^3+6t^2-10t+2`

`p(t)=-t^3+6t^2-10t+2`

The Cayley-Hamilton theorem yields that
`O = p(A)=-A^3+6A^2-10A+2I`

Rearranging terms, we have
`:. 2I = A^3-6A^2+10A`

`:. 2I = A(A^2-6A+10I)`

`:. A^-1 = 1/2(A^2-6A+10I)`

Now, first we find `A^2-6A+10I`

`A^2`

=

`A×A`

=

`3` `1` `1` `-1` `2` `1` `1` `1` `1`

×

`3` `1` `1` `-1` `2` `1` `1` `1` `1`

=	`3×3+1×-1+1×1` `3×1+1×2+1×1` `3×1+1×1+1×1` `-1×3+2×-1+1×1` `-1×1+2×2+1×1` `-1×1+2×1+1×1` `1×3+1×-1+1×1` `1×1+1×2+1×1` `1×1+1×1+1×1`

=	`9-1+1` `3+2+1` `3+1+1` `-3-2+1` `-1+4+1` `-1+2+1` `3-1+1` `1+2+1` `1+1+1`

=	`9` `6` `5` `-4` `4` `2` `3` `4` `3`

`A^2`

=

`3` `1` `1` `-1` `2` `1` `1` `1` `1` 2

=

`9` `6` `5` `-4` `4` `2` `3` `4` `3`

`6 × A`

=

`6`

×

`3` `1` `1` `-1` `2` `1` `1` `1` `1`

=

`18` `6` `6` `-6` `12` `6` `6` `6` `6`

`A^2 - 6 × A`

=

`9` `6` `5` `-4` `4` `2` `3` `4` `3`

-

`18` `6` `6` `-6` `12` `6` `6` `6` `6`

=

`9+18` `6+6` `5+6` `-4-6` `4+12` `2+6` `3+6` `4+6` `3+6`

=

`-9` `0` `-1` `2` `-8` `-4` `-3` `-2` `-3`

`10 × I`

=

`10`

×

`1` `0` `0` `0` `1` `0` `0` `0` `1`

=

`10` `0` `0` `0` `10` `0` `0` `0` `10`

`A^2 - 6 × A + 10 × I`

=

`-9` `0` `-1` `2` `-8` `-4` `-3` `-2` `-3`

+

`10` `0` `0` `0` `10` `0` `0` `0` `10`

=

`-9+10` `0+0` `-1+0` `2+0` `-8+10` `-4+0` `-3+0` `-2+0` `-3+10`

=

`1` `0` `-1` `2` `2` `-4` `-3` `-2` `7`

Now, `A^-1 = 1/2(A^2-6A+10I)`

`:. A^-1 = `

`1/(2)`

`1` `0` `-1` `2` `2` `-4` `-3` `-2` `7`