JNTU B.Tech I Year MATHEMATICS-I Previous Question Paper (September/October - 2021 - R13)

We are asked to find the non-trivial solution of the system of linear equations:

This system can be represented in matrix form as Ax = 0, where:

A non-trivial solution exists if and only if the determinant of the coefficient matrix A is zero. Let's calculate the determinant of A:

Since the determinant is zero, non-trivial solutions exist. We can use Gaussian elimination to solve the system. The augmented matrix is:

Swap R1 and R3 to get a leading 1:

Perform R2 = R2 - 3*R1 and R3 = R3 - 2*R1:

Simplify R2 by dividing by 14:

Perform R3 = R3 - 7*R2:

The system is reduced to:

From the second equation, we get y = z. Let z = k, where k is an arbitrary constant. Then y = k.

Substitute y = k and z = k into the first equation:

Thus, the non-trivial solution is x = -k, y = k, z = k. This can be written in vector form as:

where k is any non-zero real number.

The given quadratic form is Q = x² + y² + z² - 2xy - 2yz - 2zx. The corresponding symmetric matrix A is:

To find the canonical form, we determine the eigenvalues by solving the characteristic equation |A - λI| = 0:

Solving the characteristic equation gives the eigenvalues λ₁ = 2, λ₂ = 2, and λ₃ = -1. The canonical form is given by λ₁X² + λ₂Y² + λ₃Z², which results in:

The orthogonal reduction process involves finding the normalized eigenvectors corresponding to these eigenvalues to form an orthogonal matrix P such that P^TAP = D, where D is the diagonal matrix of eigenvalues.