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, if solutions are helpful to you.
 
Copyright © 2019. All rights reserved. Terms, Privacy