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3. Cayley Hamilton method example
( Enter your problem )
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- 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]]`
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Other related methods
- Adjoint method
- Gauss-Jordan Elimination method
- Cayley Hamilton method
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1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
(Previous example) | 3. Example `[[2,3],[4,10]]`
(Next example) |
2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
Find Inverse of matrix using Cayley Hamilton method
`A=[[7,2,-2],[-6,-1,2],[6,2,-1]]`
Solution:
To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial p(t) of the matrix A.
`|A-tI|`
| = | | `(7-t)` | `2` | `-2` | | | `-6` | `(-1-t)` | `2` | | | `6` | `2` | `(-1-t)` | |
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`=(7-t)((-1-t) × (-1-t) - 2 × 2)-2((-6) × (-1-t) - 2 × 6)+(-2)((-6) × 2 - (-1-t) × 6)`
`=(7-t)((1+2t+t^2)-4)-2((6+6t)-12)-2((-12)-(-6-6t))`
`=(7-t)(-3+2t+t^2)-2(-6+6t)-2(-6+6t)`
`= (-21+17t+5t^2-t^3)-(-12+12t)-(-12+12t)`
`=-t^3+5t^2-7t+3`
`p(t)=-t^3+5t^2-7t+3`
The Cayley-Hamilton theorem yields that
`O = p(A)=-A^3+5A^2-7A+3I`
Rearranging terms, we have
`:. 3I = A^3-5A^2+7A`
`:. 3I = A(A^2-5A+7I)`
`:. A^-1 = 1/3(A^2-5A+7I)`
Now, first we find `A^2-5A+7I`
| `A^2` | = | `A×A` | = | | `7` | `2` | `-2` | | | `-6` | `-1` | `2` | | | `6` | `2` | `-1` | |
| × | | `7` | `2` | `-2` | | | `-6` | `-1` | `2` | | | `6` | `2` | `-1` | |
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| = | | `7×7+2×-6-2×6` | `7×2+2×-1-2×2` | `7×-2+2×2-2×-1` | | | `-6×7-1×-6+2×6` | `-6×2-1×-1+2×2` | `-6×-2-1×2+2×-1` | | | `6×7+2×-6-1×6` | `6×2+2×-1-1×2` | `6×-2+2×2-1×-1` | |
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| = | | `49-12-12` | `14-2-4` | `-14+4+2` | | | `-42+6+12` | `-12+1+4` | `12-2-2` | | | `42-12-6` | `12-2-2` | `-12+4+1` | |
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| = | | `25` | `8` | `-8` | | | `-24` | `-7` | `8` | | | `24` | `8` | `-7` | |
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| `A^2` | = | | `7` | `2` | `-2` | | | `-6` | `-1` | `2` | | | `6` | `2` | `-1` | |
| 2 |
| = | | `25` | `8` | `-8` | | | `-24` | `-7` | `8` | | | `24` | `8` | `-7` | |
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| `5 × A` | = | `5` | × | | `7` | `2` | `-2` | | | `-6` | `-1` | `2` | | | `6` | `2` | `-1` | |
| = | | `35` | `10` | `-10` | | | `-30` | `-5` | `10` | | | `30` | `10` | `-5` | |
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| `A^2 - 5 × A` | = | | `25` | `8` | `-8` | | | `-24` | `-7` | `8` | | | `24` | `8` | `-7` | |
| - | | `35` | `10` | `-10` | | | `-30` | `-5` | `10` | | | `30` | `10` | `-5` | |
| = | | `25+35` | `8+10` | `-8-10` | | | `-24-30` | `-7-5` | `8+10` | | | `24+30` | `8+10` | `-7-5` | |
| = | | `-10` | `-2` | `2` | | | `6` | `-2` | `-2` | | | `-6` | `-2` | `-2` | |
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| `7 × I` | = | `7` | × | | `1` | `0` | `0` | | | `0` | `1` | `0` | | | `0` | `0` | `1` | |
| = | | `7` | `0` | `0` | | | `0` | `7` | `0` | | | `0` | `0` | `7` | |
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| `A^2 - 5 × A + 7 × I` | = | | `-10` | `-2` | `2` | | | `6` | `-2` | `-2` | | | `-6` | `-2` | `-2` | |
| + | | `7` | `0` | `0` | | | `0` | `7` | `0` | | | `0` | `0` | `7` | |
| = | | `-10+7` | `-2+0` | `2+0` | | | `6+0` | `-2+7` | `-2+0` | | | `-6+0` | `-2+0` | `-2+7` | |
| = | | `-3` | `-2` | `2` | | | `6` | `5` | `-2` | | | `-6` | `-2` | `5` | |
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Now, `A^-1 = 1/3(A^2-5A+7I)`
| `:. A^-1 = ` | `1/(3)` | | `-3` | `-2` | `2` | | | `6` | `5` | `-2` | | | `-6` | `-2` | `5` | |
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This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
1. Example `[[3,1,1],[-1,2,1],[1,1,1]]`
(Previous example) | 3. Example `[[2,3],[4,10]]`
(Next example) |
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