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3. Cayley Hamilton method example ( Enter your problem )
( Enter your problem )
  1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
  2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
  3. Example `[[2,3],[4,10]]`
  4. Example `[[5,1],[4,2]]`
 		Other related methods
  1. Adjoint method
  2. Gauss-Jordan Elimination method
  3. Cayley Hamilton method
2. Gauss-Jordan Elimination method
(Previous method)		2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
(Next example)

1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`


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`


This material is intended as a summary. Use your textbook for detail explanation.
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2. Gauss-Jordan Elimination method
(Previous method)		2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
(Next example)


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