Math, asked by navyasrirangu134, 1 month ago

find the matrix of the quadratic form x^2+5y^2+z^2+2xy+2yz+6zx and it's nature

Answers

Answered by Mithalesh1602398
0

Answer:

The matrix of the quadratic form is:

[1 1 3; 1 5 1; 3 1 1]

Step-by-step explanation:

The quadratic form can be written in matrix form as:

Q = [x y z] [1 1 3; 1 5 1; 3 1 1] [x; y; z]

Thus, the matrix of the quadratic form is:

[1 1 3; 1 5 1; 3 1 1]

To determine the nature of the quadratic form, we can find its eigenvalues and eigenvectors.

The characteristic equation is:

det([1-lambda 1 3; 1 5-lambda 1; 3 1 1-lambda]) = 0

Expanding the determinant along the first row, we get:

(1-lambda) [(5-lambda)(1-lambda) - 1] - (1) [1(1-lambda) - 3(1)] + (3) [1-5(1)] = 0

Simplifying, we get:

lambda^3 - 7lambda^2 + 12lambda - 4 = 0

The roots of this cubic equation are lambda = 4, 1, 1

To find the corresponding eigenvectors, we can solve the systems of linear equations:

[1 1 3; 1 5 1; 3 1 1] [x; y; z] = 4 [x; y; z]

[1 1 3; 1 5 1; 3 1 1] [x; y; z] = 1 [x; y; z]

[1 1 3; 1 5 1; 3 1 1] [x; y; z] = 1 [x; y; z]

For lambda = 4, we get the system:

-3x + y + 3z = 0

x - 3y + z = 0

3x + y - 3z = 0

Solving this system, we get the eigenvector:

v1 = [1; -1; 1]

For lambda = 1, we get the system:

x + y + 3z = 0

x - y + z = 0

3x + y + z = 0

Solving this system, we get the eigenvectors:

v2 = [-1; 0; 1]

v3 = [1; -2; 1]

Since all three eigenvalues are positive, the quadratic form is positive definite.

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