Inverse of matrix using Cayley Hamilton method Example [[3,1,1],[-1,2,1],[1,1,1]]

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Home > Matrix & Vector calculators > Inverse of matrix using Cayley Hamilton method example

3. Cayley Hamilton method example ( Enter your problem )

( Enter your problem )

Example `[[3,1,1],[-1,2,1],[1,1,1]]`
Example `[[2,3,1],[0,5,6],[1,1,2]]`
Example `[[2,3],[4,10]]`
Example `[[5,1],[4,2]]`

Other related methods

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]]`

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`

`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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