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Home > Matrix & Vector calculators > is Derogatory Matrix example
23. is Derogatory Matrix example ( Enter your problem )
( Enter your problem )
  1. Definition & Examples
  2. Example-2
 		Other related methods
  1. is Row Matrix
  2. is Column Matrix
  3. is Square Matrix
  4. is Horizontal Matrix
  5. is Vertical Matrix
  6. is Diagonal Matrix
  7. is Identity Matrix
  8. is Scalar Matrix
  9. is Null Matrix
  10. is Lower Triangle Matrix
  11. is Upper Triangle Matrix
  12. is Orthogonal Matrix
  13. is Singular Matrix
  14. is Nonsingular Matrix
  15. is Symmetric Matrix
  16. is Skew Symmetric Matrix
  17. is Nilpotent Matrix
  18. is Involutary Matrix
  19. is Idempotent Matrix
  20. is Periodic Matrix
  21. is Positive Definite Matrix
  22. is Negative Definite Matrix
  23. is Derogatory Matrix
  24. is Diagonally Dominant Matrix
  25. is Strictly Diagonally Dominant Matrix
  26. Auto detect the matrix type
1. Definition & Examples
(Previous example)		24. is Diagonally Dominant Matrix
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2. Example-2


1. is Derogatory Matrix ?
`[[7,4,-1],[4,7,-1],[-4,-4,4]]`

Solution:
`A` = 
`7``4``-1`
`4``7``-1`
`-4``-4``4`


A matrix is derogatory if it's eigenvalues are repeated.

`|A-lamdaI|=0`

 `(7-lamda)`  `4`  `-1` 
 `4`  `(7-lamda)`  `-1` 
 `-4`  `-4`  `(4-lamda)` 
 = 0


`:.(7-lamda)((7-lamda) × (4-lamda) - (-1) × (-4))-4(4 × (4-lamda) - (-1) × (-4))+(-1)(4 × (-4) - (7-lamda) × (-4))=0`

`:.(7-lamda)((28-11lamda+lamda^2)-4)-4((16-4lamda)-4)-1((-16)-(-28+4lamda))=0`

`:.(7-lamda)(24-11lamda+lamda^2)-4(12-4lamda)-1(12-4lamda)=0`

`:. (168-101lamda+18lamda^2-lamda^3)-(48-16lamda)-(12-4lamda)=0`

`:.(-lamda^3+18lamda^2-81lamda+108)=0`

`:.-(lamda^3-18lamda^2+81lamda-108)=0`

`:.(lamda^3-18lamda^2+81lamda-108)=0 `

Roots can be found using newton raphson method
Newton Raphson method for `x^3-18x^2+81x-108=0`


Here `x^3-18x^2+81x-108=0`

Let `f(x) = x^3-18x^2+81x-108`

`:. f'(x) = 3x^2-36x+81`

`x_0 = 2`


`1^(st)` iteration :

`f(x_0)=f(2)=2^3-18 xx 2^2+81 xx 2-108=-10`

`f'(x_0)=f'(2)=3 xx 2^2-36 xx 2+81=21`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=2 - (-10)/(21)`

`x_1=2.47619048`


`2^(nd)` iteration :

`f(x_1)=f(2.47619048)=2.47619048^3-18 xx 2.47619048^2+81 xx 2.47619048-108=-2.61310874`

`f'(x_1)=f'(2.47619048)=3 xx 2.47619048^2-36 xx 2.47619048+81=10.25170068`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=2.47619048 - (-2.61310874)/(10.25170068)`

`x_2=2.73108562`


`3^(rd)` iteration :

`f(x_2)=f(2.73108562)=2.73108562^3-18 xx 2.73108562^2+81 xx 2.73108562-108=-0.67028101`

`f'(x_2)=f'(2.73108562)=3 xx 2.73108562^2-36 xx 2.73108562+81=5.05740364`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=2.73108562 - (-0.67028101)/(5.05740364)`

`x_3=2.86362023`


`4^(th)` iteration :

`f(x_3)=f(2.86362023)=2.86362023^3-18 xx 2.86362023^2+81 xx 2.86362023-108=-0.16993156`

`f'(x_3)=f'(2.86362023)=3 xx 2.86362023^2-36 xx 2.86362023+81=2.51063417`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4=2.86362023 - (-0.16993156)/(2.51063417)`

`x_4=2.93130495`


`5^(th)` iteration :

`f(x_4)=f(2.93130495)=2.93130495^3-18 xx 2.93130495^2+81 xx 2.93130495-108=-0.04279527`

`f'(x_4)=f'(2.93130495)=3 xx 2.93130495^2-36 xx 2.93130495+81=1.25066799`

`x_5 = x_4 - f(x_4)/(f'(x_4))`

`x_5=2.93130495 - (-0.04279527)/(1.25066799)`

`x_5=2.96552287`


`6^(th)` iteration :

`f(x_5)=f(2.96552287)=2.96552287^3-18 xx 2.96552287^2+81 xx 2.96552287-108=-0.01073903`

`f'(x_5)=f'(2.96552287)=3 xx 2.96552287^2-36 xx 2.96552287+81=0.62415429`

`x_6 = x_5 - f(x_5)/(f'(x_5))`

`x_6=2.96552287 - (-0.01073903)/(0.62415429)`

`x_6=2.98272861`


`7^(th)` iteration :

`f(x_6)=f(2.98272861)=2.98272861^3-18 xx 2.98272861^2+81 xx 2.98272861-108=-0.00268986`

`f'(x_6)=f'(2.98272861)=3 xx 2.98272861^2-36 xx 2.98272861+81=0.31177998`

`x_7 = x_6 - f(x_6)/(f'(x_6))`

`x_7=2.98272861 - (-0.00268986)/(0.31177998)`

`x_7=2.99135604`


`8^(th)` iteration :

`f(x_7)=f(2.99135604)=2.99135604^3-18 xx 2.99135604^2+81 xx 2.99135604-108=-0.00067311`

`f'(x_7)=f'(2.99135604)=3 xx 2.99135604^2-36 xx 2.99135604+81=0.15581542`

`x_8 = x_7 - f(x_7)/(f'(x_7))`

`x_8=2.99135604 - (-0.00067311)/(0.15581542)`

`x_8=2.99567595`


`9^(th)` iteration :

`f(x_8)=f(2.99567595)=2.99567595^3-18 xx 2.99567595^2+81 xx 2.99567595-108=-0.00016836`

`f'(x_8)=f'(2.99567595)=3 xx 2.99567595^2-36 xx 2.99567595+81=0.07788903`

`x_9 = x_8 - f(x_8)/(f'(x_8))`

`x_9=2.99567595 - (-0.00016836)/(0.07788903)`

`x_9=2.99783745`


`10^(th)` iteration :

`f(x_9)=f(2.99783745)=2.99783745^3-18 xx 2.99783745^2+81 xx 2.99783745-108=-0.0000421`

`f'(x_9)=f'(2.99783745)=3 xx 2.99783745^2-36 xx 2.99783745+81=0.03893984`

`x_10 = x_9 - f(x_9)/(f'(x_9))`

`x_10=2.99783745 - (-0.0000421)/(0.03893984)`

`x_10=2.9989186`


`11^(th)` iteration :

`f(x_10)=f(2.9989186)=2.9989186^3-18 xx 2.9989186^2+81 xx 2.9989186-108=-0.00001053`

`f'(x_10)=f'(2.9989186)=3 xx 2.9989186^2-36 xx 2.9989186+81=0.01946875`

`x_11 = x_10 - f(x_10)/(f'(x_10))`

`x_11=2.9989186 - (-0.00001053)/(0.01946875)`

`x_11=2.99945927`


`12^(th)` iteration :

`f(x_11)=f(2.99945927)=2.99945927^3-18 xx 2.99945927^2+81 xx 2.99945927-108=-0.00000263`

`f'(x_11)=f'(2.99945927)=3 xx 2.99945927^2-36 xx 2.99945927+81=0.00973408`

`x_12 = x_11 - f(x_11)/(f'(x_11))`

`x_12=2.99945927 - (-0.00000263)/(0.00973408)`

`x_12=2.99972963`


`13^(th)` iteration :

`f(x_12)=f(2.99972963)=2.99972963^3-18 xx 2.99972963^2+81 xx 2.99972963-108=-0.00000066`

`f'(x_12)=f'(2.99972963)=3 xx 2.99972963^2-36 xx 2.99972963+81=0.00486697`

`x_13 = x_12 - f(x_12)/(f'(x_12))`

`x_13=2.99972963 - (-0.00000066)/(0.00486697)`

`x_13=2.99986481`


`14^(th)` iteration :

`f(x_13)=f(2.99986481)=2.99986481^3-18 xx 2.99986481^2+81 xx 2.99986481-108=-0.00000016`

`f'(x_13)=f'(2.99986481)=3 xx 2.99986481^2-36 xx 2.99986481+81=0.00243347`

`x_14 = x_13 - f(x_13)/(f'(x_13))`

`x_14=2.99986481 - (-0.00000016)/(0.00243347)`

`x_14=2.9999324`


`15^(th)` iteration :

`f(x_14)=f(2.9999324)=2.9999324^3-18 xx 2.9999324^2+81 xx 2.9999324-108=-0.00000004`

`f'(x_14)=f'(2.9999324)=3 xx 2.9999324^2-36 xx 2.9999324+81=0.00121673`

`x_15 = x_14 - f(x_14)/(f'(x_14))`

`x_15=2.9999324 - (-0.00000004)/(0.00121673)`

`x_15=2.9999662`


Approximate root of the equation `x^3-18x^2+81x-108=0` using Newton Raphson method is `2.9999662` (After 15 iterations)

`n``x_0``f(x_0)``f'(x_0)``x_1`Update
12-10212.47619048`x_0 = x_1`
22.47619048-2.6131087410.251700682.73108562`x_0 = x_1`
32.73108562-0.670281015.057403642.86362023`x_0 = x_1`
42.86362023-0.169931562.510634172.93130495`x_0 = x_1`
52.93130495-0.042795271.250667992.96552287`x_0 = x_1`
62.96552287-0.010739030.624154292.98272861`x_0 = x_1`
72.98272861-0.002689860.311779982.99135604`x_0 = x_1`
82.99135604-0.000673110.155815422.99567595`x_0 = x_1`
92.99567595-0.000168360.077889032.99783745`x_0 = x_1`
102.99783745-0.00004210.038939842.9989186`x_0 = x_1`
112.9989186-0.000010530.019468752.99945927`x_0 = x_1`
122.99945927-0.000002630.009734082.99972963`x_0 = x_1`
132.99972963-0.000000660.004866972.99986481`x_0 = x_1`
142.99986481-0.000000160.002433472.9999324`x_0 = x_1`
152.9999324-0.000000040.001216732.9999662`x_0 = x_1`


`:. `x=2.9999662


Now, using long division `(x^3-18x^2+81x-108)/(x-2.9999662)=x^2-15.0000338x+36.00040557`


Now, `x^2-15.0000338x+36.00040557=0`

`:. x=3.0000338` and `x=12`


`:.` The eigenvalues of the matrix A are given by `lamda=2.9999662,12`

`:.` The eigenvalues of the matrix A are `3,3,12`

Here eigenvalue `3` is repeated, so matrix is derogatory matrix.

This material is intended as a summary. Use your textbook for detail explanation.
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1. Definition & Examples
(Previous example)		24. is Diagonally Dominant Matrix
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