Question

eigenvalues of the matrix A = [1 3 0, 3 ?2 ?1, 0 ?1 1] are 1, ?4 and 3. express the equation of the surface x^2 ? 2y^2 + z^2 + 6xy ? 2yz = 16. How should i determine the order of the coefficient in the form X^2/A+Y^2/B+Z^2/C=1?

Answer 1

Given matrix is A = and eigenvalues of the matrix A are 1, -4 and 3.

For = 1 : AX = X

i.e., (A-I)X = 0

i.e., =

i.e., 3y = 0

3x-3y-z = 0

-y = 0

i.e., y = 0

z = 3x

Let us take x = k. Then, z = 3k.

Therefore, the eigenvectors corresponding to the eigenvalues 1 are ,i.e., , where k is non-zero real number.

For = 1 : AX = -4X

i.e., (A+4I)X = 0

i.e., =

i.e., 5x+3y = 0

3x+2y-z = 0

-y+5z = 0

i.e., y = 5z

x = -3z

Let us take z = k. Then, x = -3k and y = 5k.

Therefore, the eigenvectors corresponding to the eigenvalue -4 are ,i.e., , where k is non-zero real number.

For = 1 : AX = 3X

i.e., (A-3I)X = 0

i.e., =

i.e., -2x+3y = 0

3x-5y-z = 0

-y-2z = 0

i.e., x = -3z

y = -2z

Let us take z = k. Then, x = -3k and y = -2k.

Therefore, the eigenvectors corresponding to the eigenvalue 3 are ,i.e., , where k is non-zero real number.

Here, the set is an orthogonal set. The orthonormal set of eigenvectors is

Let P = .Then, P is an orthogonal matrix.

Now, P^TAP =

=

=

Here, the given equation is X^TAX = 16I.

By the transformation X = PX', where X' = the equation reduces to X'^T(P^TAP)X = 16I.

Now, X'^T(P^TAP)X = 16I gives, x'²-4y'²+3z'² = 16

i.e.,

i.e.,

Therefore, the coefficients of x'², y'², z'² are , , .