Home > Matrix Algebra calculators > Inverse of matrix using Cayley Hamilton method calculator

Solve any problem
(step by step solutions)
Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
Find Cayley Hamilton inverse [[10,-9,-12],[7,-12,11],[-10,10,3]]

Solution:
Your problem `->` Cayley Hamilton inverse [[10,-9,-12],[7,-12,11],[-10,10,3]]


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

 = 
 `(10-t)`  `-9`  `-12` 
 `7`  `(-12-t)`  `11` 
 `-10`  `10`  `(3-t)` 


`=(10-t)((-12-t) × (3-t) - 11 × 10)-(-9)(7 × (3-t) - 11 × (-10))+(-12)(7 × 10 - (-12-t) × (-10))`

`=(10-t)((-36+9t+t^2)-110)+9((21-7t)-(-110))-12(70-(120+10t))`

`=(10-t)(-146+9t+t^2)+9(131-7t)-12(-50-10t)`

`= (-1460+236t+t^2-t^3)+(1179-63t)-(-600-120t)`

`=-t^3+t^2+293t+319`

`p(t)=-t^3+t^2+293t+319`

The Cayley-Hamilton theorem yields that
`O = p(A)=-A^3+A^2+293A+319I`

Rearranging terms, we have
`:. 319I = A^3-A^2-293A`

`:. 319I = A(A^2-A-293I)`

`:. A^-1 = 1/319(A^2-A-293I)`

Now, first we find `A^2-A-293I`

`A^2`=`A×A`=
`10``-9``-12`
`7``-12``11`
`-10``10``3`
×
`10``-9``-12`
`7``-12``11`
`-10``10``3`


=
`10×10-9×7-12×-10``10×-9-9×-12-12×10``10×-12-9×11-12×3`
`7×10-12×7+11×-10``7×-9-12×-12+11×10``7×-12-12×11+11×3`
`-10×10+10×7+3×-10``-10×-9+10×-12+3×10``-10×-12+10×11+3×3`


=
`100-63+120``-90+108-120``-120-99-36`
`70-84-110``-63+144+110``-84-132+33`
`-100+70-30``90-120+30``120+110+9`


=
`157``-102``-255`
`-124``191``-183`
`-60``0``239`


`A^2` = 
`10``-9``-12`
`7``-12``11`
`-10``10``3`
2
 = 
`157``-102``-255`
`-124``191``-183`
`-60``0``239`


`A^2 - A` = 
`157``-102``-255`
`-124``191``-183`
`-60``0``239`
 - 
`10``-9``-12`
`7``-12``11`
`-10``10``3`
 = 
`157+10``-102-9``-255-12`
`-124+7``191-12``-183+11`
`-60-10``0+10``239+3`
 = 
`147``-93``-243`
`-131``203``-194`
`-50``-10``236`


`293 × I` = `293` × 
`1``0``0`
`0``1``0`
`0``0``1`
 = 
`293``0``0`
`0``293``0`
`0``0``293`


`A^2 - A - 293 × I` = 
`147``-93``-243`
`-131``203``-194`
`-50``-10``236`
 - 
`293``0``0`
`0``293``0`
`0``0``293`
 = 
`147+293``-93+0``-243+0`
`-131+0``203+293``-194+0`
`-50+0``-10+0``236+293`
 = 
`-146``-93``-243`
`-131``-90``-194`
`-50``-10``-57`


Now, `A^-1 = 1/319(A^2-A-293I)`

`:. A^-1 = ``1/(319)`
`-146``-93``-243`
`-131``-90``-194`
`-50``-10``-57`







Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me
  
 

Share with your friends
 
Copyright © 2018. All rights reserved. Terms, Privacy