is Derogatory Matrix 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
1	2	-10	21	2.47619048	`x_0 = x_1`
2	2.47619048	-2.61310874	10.25170068	2.73108562	`x_0 = x_1`
3	2.73108562	-0.67028101	5.05740364	2.86362023	`x_0 = x_1`
4	2.86362023	-0.16993156	2.51063417	2.93130495	`x_0 = x_1`
5	2.93130495	-0.04279527	1.25066799	2.96552287	`x_0 = x_1`
6	2.96552287	-0.01073903	0.62415429	2.98272861	`x_0 = x_1`
7	2.98272861	-0.00268986	0.31177998	2.99135604	`x_0 = x_1`
8	2.99135604	-0.00067311	0.15581542	2.99567595	`x_0 = x_1`
9	2.99567595	-0.00016836	0.07788903	2.99783745	`x_0 = x_1`
10	2.99783745	-0.0000421	0.03893984	2.9989186	`x_0 = x_1`
11	2.9989186	-0.00001053	0.01946875	2.99945927	`x_0 = x_1`
12	2.99945927	-0.00000263	0.00973408	2.99972963	`x_0 = x_1`
13	2.99972963	-0.00000066	0.00486697	2.99986481	`x_0 = x_1`
14	2.99986481	-0.00000016	0.00243347	2.9999324	`x_0 = x_1`
15	2.9999324	-0.00000004	0.00121673	2.9999662	`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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2. Example-2