1. is Positive Definite Matrix ?
`[[25,15,-5],[15,18,0],[-5,0,11]]`Solution:| `A` | = | | `25` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
|
A matrix is positive definite if it's symmetric and all its pivots are positive.
Test method 1: Existence of all positive Pivots.First apply Gaussian Elimination method to find Pivots
| `A` | = | | `25` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
|
`R_2 larr R_2-0.6xx R_1`
| = | | `25` | `15` | `-5` | | | `0` | `9` | `3` | | | `-5` | `0` | `11` | |
|
`R_3 larr R_3+0.2xx R_1`
| = | | `25` | `15` | `-5` | | | `0` | `9` | `3` | | | `0` | `3` | `10` | |
|
`R_3 larr R_3-0.3333xx R_2`
| = | | `25` | `15` | `-5` | | | `0` | `9` | `3` | | | `0` | `0` | `9` | |
|
Pivots are the first non-zero element in each row of this eliminated matrix.
`:.` Pivots are `25,9,9`
Here all pivots are positive, so matrix is positive definite.
A matrix is positive definite if Determinants of all upper-left sub-matrices are positive.
Test method 2: Determinants of all upper-left sub-matrices are positive.| `A` | = | | `25` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
|
| `25` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
| `=2025` |
Determinants are `25,225,2025`
Here all determinants are positive, so matrix is positive definite.
A matrix is positive definite if it's symmetric and all its eigenvalues are positive.
Test method 3: All positive eigen values.`|A-lamdaI|=0`
| `(25-lamda)` | `15` | `-5` | | | `15` | `(18-lamda)` | `0` | | | `-5` | `0` | `(11-lamda)` | |
| = 0 |
`:.(25-lamda)((18-lamda) × (11-lamda) - 0 × 0)-15(15 × (11-lamda) - 0 × (-5))+(-5)(15 × 0 - (18-lamda) × (-5))=0`
`:.(25-lamda)((198-29lamda+lamda^2)-0)-15((165-15lamda)-0)-5(0-(-90+5lamda))=0`
`:.(25-lamda)(198-29lamda+lamda^2)-15(165-15lamda)-5(90-5lamda)=0`
`:. (4950-923lamda+54lamda^2-lamda^3)-(2475-225lamda)-(450-25lamda)=0`
`:.(-lamda^3+54lamda^2-673lamda+2025)=0`
`:.-(lamda^3-54lamda^2+673lamda-2025)=0`
`:.(lamda^3-54lamda^2+673lamda-2025)=0 `
Roots can be found using newton raphson method
Newton Raphson method for `x^3-54x^2+673x-2025=0`
Here `x^3-54x^2+673x-2025=0`
Let `f(x) = x^3-54x^2+673x-2025`
`:. f'(x) = 3x^2-108x+673`
`x_0 = 4`
`1^(st)` iteration :`f(x_0)=f(4)=4^3-54 xx 4^2+673 xx 4-2025=-133`
`f'(x_0)=f'(4)=3 xx 4^2-108 xx 4+673=289`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=4 - (-133)/(289)`
`x_1=4.46020761`
`2^(nd)` iteration :`f(x_1)=f(4.46020761)=4.46020761^3-54 xx 4.46020761^2+673 xx 4.46020761-2025=-8.7977561`
`f'(x_1)=f'(4.46020761)=3 xx 4.46020761^2-108 xx 4.46020761+673=250.97793369`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=4.46020761 - (-8.7977561)/(250.97793369)`
`x_2=4.49526152`
`3^(rd)` iteration :`f(x_2)=f(4.49526152)=4.49526152^3-54 xx 4.49526152^2+673 xx 4.49526152-2025=-0.04986905`
`f'(x_2)=f'(4.49526152)=3 xx 4.49526152^2-108 xx 4.49526152+673=248.13388462`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3=4.49526152 - (-0.04986905)/(248.13388462)`
`x_3=4.49546249`
`4^(th)` iteration :`f(x_3)=f(4.49546249)=4.49546249^3-54 xx 4.49546249^2+673 xx 4.49546249-2025=-0.00000164`
`f'(x_3)=f'(4.49546249)=3 xx 4.49546249^2-108 xx 4.49546249+673=248.11759994`
`x_4 = x_3 - f(x_3)/(f'(x_3))`
`x_4=4.49546249 - (-0.00000164)/(248.11759994)`
`x_4=4.4954625`
Approximate root of the equation `x^3-54x^2+673x-2025=0` using Newton Raphson method is `4.4954625` (After 4 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 4 | -133 | 289 | 4.46020761 | `x_0 = x_1` |
| 2 | 4.46020761 | -8.7977561 | 250.97793369 | 4.49526152 | `x_0 = x_1` |
| 3 | 4.49526152 | -0.04986905 | 248.13388462 | 4.49546249 | `x_0 = x_1` |
| 4 | 4.49546249 | -0.00000164 | 248.11759994 | 4.4954625 | `x_0 = x_1` |
Now, using long division `(x^3-54x^2+673x-2025)/(x-4.4954625)=x^2-49.5045375x+450.45420817`
Now, `x^2-49.5045375x+450.45420817=0`
`:. x=12.01568365` and `x=37.48885386`
`:.` The eigenvalues of the matrix A are given by `lamda=4.4954625,12.01568365,37.48885386`
Here all determinants are positive, so matrix is positive definite.
