JNTU B.Tech I Year MATHEMATICS-I Previous Question Paper (July - 2021 - R15)

Let the given matrix be A. To find the rank, we perform row operations to reduce it to row-echelon form:

Applying R2 = R2 - R1:

Applying R4 = R4 - R3:

For the rank of the matrix to be 3, the number of non-zero rows in the row-echelon form must be 3. This occurs if the last row becomes a zero row. Therefore, we set the pivot element of the fourth row to zero:

Solving this gives k = 1.

The system of equations is homogeneous, represented as AX = 0, where:

For a non-trivial solution to exist, the determinant of the coefficient matrix must be zero (|A| = 0):

Substituting a = -3 into the system:

Subtracting the first equation from the second: 4y - 4z = 0, which implies y = z. Substituting y = z into the first equation: x + z + z = 0, so x = -2z. Letting z = k, the non-trivial solution is x = -2k, y = k, z = k for any non-zero constant k.