2. is Positive Definite Matrix ?
`[[2,15,-5],[15,18,0],[-5,0,11]]`
Solution:
| `A` | = | | `2` | `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` | = | | `2` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
|
`R_2 larr R_2-7.5xx R_1`
| = | | `2` | `15` | `-5` | | | `0` | `-94.5` | `37.5` | | | `-5` | `0` | `11` | |
|
`R_3 larr R_3+2.5xx R_1`
| = | | `2` | `15` | `-5` | | | `0` | `-94.5` | `37.5` | | | `0` | `37.5` | `-1.5` | |
|
`R_3 larr R_3+0.3968xx R_2`
| = | | `2` | `15` | `-5` | | | `0` | `-94.5` | `37.5` | | | `0` | `0` | `13.381` | |
|
Pivots are the first non-zero element in each row of this eliminated matrix.
`:.` Pivots are `2,-189/2,281/21`
Here `-94.5` is not positive, so matrix is not 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` | = | | `2` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
|
| `2` | `15` | `-5` | | | `15` | `18` | `0` | | | `-5` | `0` | `11` | |
| `=-2529` |
Determinants are `2,-189,-2529`
Here `-189` is not positive, so matrix is not 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`
| `(2-lamda)` | `15` | `-5` | | | `15` | `(18-lamda)` | `0` | | | `-5` | `0` | `(11-lamda)` | |
| = 0 |
`:.(2-lamda)((18-lamda) × (11-lamda) - 0 × 0)-15(15 × (11-lamda) - 0 × (-5))+(-5)(15 × 0 - (18-lamda) × (-5))=0`
`:.(2-lamda)((198-29lamda+lamda^2)-0)-15((165-15lamda)-0)-5(0-(-90+5lamda))=0`
`:.(2-lamda)(198-29lamda+lamda^2)-15(165-15lamda)-5(90-5lamda)=0`
`:. (396-256lamda+31lamda^2-lamda^3)-(2475-225lamda)-(450-25lamda)=0`
`:.(-lamda^3+31lamda^2-6lamda-2529)=0`
`:.-(lamda^3-31lamda^2+6lamda+2529)=0`
`:.(lamda^3-31lamda^2+6lamda+2529)=0 `
Roots can be found using newton raphson method
Newton Raphson method for `x^3-31x^2+6x+2529=0`
Here `x^3-31x^2+6x+2529=0`
Let `f(x) = x^3-31x^2+6x+2529`
`:. f'(x) = 3x^2-62x+6`
`x_0 = -8`
`1^(st)` iteration :`f(x_0)=f(-8)=-8^3-31 xx -8^2+6 xx -8+2529=-15`
`f'(x_0)=f'(-8)=3 xx -8^2-62 xx -8+6=694`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=-8 - (-15)/(694)`
`x_1=-7.97838617`
`2^(nd)` iteration :`f(x_1)=f(-7.97838617)=-7.97838617^3-31 xx -7.97838617^2+6 xx -7.97838617+2529=-0.02568358`
`f'(x_1)=f'(-7.97838617)=3 xx -7.97838617^2-62 xx -7.97838617+6=691.62387986`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=-7.97838617 - (-0.02568358)/(691.62387986)`
`x_2=-7.97834903`
Approximate root of the equation `x^3-31x^2+6x+2529=0` using Newton Raphson method is `-7.97834903` (After 2 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | -8 | -15 | 694 | -7.97838617 | `x_0 = x_1` |
| 2 | -7.97838617 | -0.02568358 | 691.62387986 | -7.97834903 | `x_0 = x_1` |
Now, using long division `(x^3-31x^2+6x+2529)/(x+7.97834903)=x^2-38.97834903x+316.98287327`
Now, `x^2-38.97834903x+316.98287327=0`
`:. x=11.56168752` and `x=27.41666151`
`:.` The eigenvalues of the matrix A are given by `lamda=-7.97834903,11.56168752,27.41666151`
Here `-7.9783` is not positive, so matrix is not positive definite.
This material is intended as a summary. Use your textbook for detail explanation.
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