JNTU B.Tech I MATHEMATICS I Previous Question Paper (July - 2021 - R13)

To find the value of k such that the rank of a matrix is 3, the matrix must be at least 3x3. Assuming the matrix is A = [[1, 2, 3], [2, 4, 6], [3, 6, k]], we observe that the second row is twice the first row, making the rank at most 2 regardless of k. If the matrix is of a higher dimension, such as a 3x4 or 4x4 matrix, we perform row reduction to echelon form. For a matrix to have rank 3, the determinant of the 3x3 submatrix must be non-zero. If the matrix is defined as A = [[1, 1, 1], [1, 2, 3], [1, 4, k]], we set the determinant to zero to find the value where rank is less than 3: 1(2k - 12) - 1(k - 3) + 1(4 - 2) = 0, which simplifies to 2k - 12 - k + 3 + 2 = 0, giving k = 7. Therefore, for the rank to be 3, k must not equal 7.

The given system of homogeneous equations is:

For a non-trivial solution to exist, the determinant of the coefficient matrix must be zero:

Expanding the determinant: 1(-5 - a) - 1(-1 - 2a) + 1(1 - 10) = 0. This simplifies to -5 - a + 1 + 2a - 9 = 0, resulting in a - 13 = 0, so a = 13. Substituting a = 13 into the system, we solve using row reduction:

• R2 = R2 - R1: 4y + 12z = 0 => y = -3z
• Substitute into R1: x + (-3z) + z = 0 => x = 2z
• Setting z = k, the non-trivial solution is x = 2k, y = -3k, z = k for any non-zero k.