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Problem: Is Positive Definite matrix [[25,15,-5],[15,18,0],[-5,0,11]] [ Calculator, Method and examples ]

Solution:
Your problem -> Is Positive Definite matrix [[25,15,-5],[15,18,0],[-5,0,11]]

A matrix is positive definite if it’s symmetric and all its pivots are positive.

A =
 25 15 -5 15 18 0 -5 0 11

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_2-3/5xx R_1

=
 25 15 -5 0 9 3 -5 0 11

R_3 larr R_3+1/5xx R_1

=
 25 15 -5 0 9 3 0 3 10

R_3 larr R_3-1/3xx 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.

Test method 2: Determinants of all upper-left sub-matrices are positive.
A =
 25 15 -5 15 18 0 -5 0 11

 25
=25

 25 15 15 18
=225

 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 = 0

1^(st) iteration :

f(x_0)=f(0)=1 xx 0^3-54 xx 0^2+673 xx 0-2025=-2025

f'(x_0)=f'(0)=3 xx 0^2-108 xx 0+673=673

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

x_1 = 0 - (-2025)/(673)

x_1 = 3.00891530460624

2^(nd) iteration :

f(x_1)=f(3.00891530460624)=1 xx 3.00891530460624^3-54 xx 3.00891530460624^2+673 xx 3.00891530460624-2025=-461.651421478971

f'(x_1)=f'(3.00891530460624)=3 xx 3.00891530460624^2-108 xx 3.00891530460624+673=375.197861033407

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

x_2 = 3.00891530460624 - (-461.651421478971)/(375.197861033407)

x_2 = 4.23933655542959

3^(rd) iteration :

f(x_2)=f(4.23933655542959)=1 xx 4.23933655542959^3-54 xx 4.23933655542959^2+673 xx 4.23933655542959-2025=-66.223869251573

f'(x_2)=f'(4.23933655542959)=3 xx 4.23933655542959^2-108 xx 4.23933655542959+673=269.067575304209

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

x_3 = 4.23933655542959 - (-66.223869251573)/(269.067575304209)

x_3 = 4.48546011445264

4^(th) iteration :

f(x_3)=f(4.48546011445264)=1 xx 4.48546011445264^3-54 xx 4.48546011445264^2+673 xx 4.48546011445264-2025=-2.48582175321596

f'(x_3)=f'(4.48546011445264)=3 xx 4.48546011445264^2-108 xx 4.48546011445264+673=248.928364954152

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

x_4 = 4.48546011445264 - (-2.48582175321596)/(248.928364954152)

x_4 = 4.49544620725357

5^(th) iteration :

f(x_4)=f(4.49544620725357)=1 xx 4.49544620725357^3-54 xx 4.49544620725357^2+673 xx 4.49544620725357-2025=-0.004042097009886

f'(x_4)=f'(4.49544620725357)=3 xx 4.49544620725357^2-108 xx 4.49544620725357+673=248.118919423546

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

x_5 = 4.49544620725357 - (-0.004042097009886)/(248.118919423546)

x_5 = 4.49546249822009

6^(th) iteration :

f(x_5)=f(4.49546249822009)=1 xx 4.49546249822009^3-54 xx 4.49546249822009^2+673 xx 4.49546249822009-2025=-0.000000010752174

f'(x_5)=f'(4.49546249822009)=3 xx 4.49546249822009^2-108 xx 4.49546249822009+673=248.11759941094

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

x_6 = 4.49546249822009 - (-0.000000010752174)/(248.11759941094)

x_6 = 4.49546249826343

Approximate root of the equation x^3-54x^2+673x-2025=0 using Newton Raphson mehtod is 4.49546249826343

 n x_0 f(x_0) f'(x_0) x_1 Update 1 0 -2025 673 3.00891530460624 x_0 = x_1 2 3.00891530460624 -461.651421478971 375.197861033407 4.23933655542959 x_0 = x_1 3 4.23933655542959 -66.223869251573 269.067575304209 4.48546011445264 x_0 = x_1 4 4.48546011445264 -2.48582175321596 248.928364954152 4.49544620725357 x_0 = x_1 5 4.49544620725357 -0.004042097009886 248.118919423546 4.49546249822009 x_0 = x_1 6 4.49546249822009 -0.000000010752174 248.11759941094 4.49546249826343 x_0 = x_1

:. x=4.49546

Now, using long division (x^3-54x^2+673x-2025)/(x-4.49546)=x^2-49.50454x+450.45421

Now, x^2-49.50454x+450.45421=0

:. x=12.01568 and x=37.48885

:. The eigenvalues of the matrix A are given by lamda=4.49546,12.01568,37.48885

Here all determinants are positive, so matrix is positive definite.

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