If x, y, z are in arithmetic progression with common difference d, x ≠ 3d, and the determinant of the matrix [ 3, 4√2, x; 4, 5√2, y; 5, k, z ] is zero, then the value of k² is :

Solution

Maths sir · 4 steps

Use the AP: R₁ + R₃ - 2R₂ kills two entries of a row, and k pops right out.

1. Use the AP. Since x, y, z are in AP, x + z = 2y. The first column 3, 4, 5 is also an AP, so 3 + 5 = 2 × 4. Apply R₁ → R₁ + R₃ - 2R₂. This row operation does not change the determinant.

   |0, k - 6√2, 0; 4, 5√2, y; 5, k, z| = 0

2. Expand along row 1. Only the middle entry survives: -(k - 6√2)(4z - 5y) = (k - 6√2)(5y - 4z). With y = x + d and z = x + 2d, we get 5y - 4z = x - 3d.
3. Use the condition. The determinant is (k - 6√2)(x - 3d) = 0. We are told x ≠ 3d, so the second factor is not zero. That forces k = 6√2.
4. So. k² = 36 × 2 = 72, which is option (a).

The idea to keep. When a column holds an AP, try R₁ + R₃ - 2R₂. It turns entries into zeros and makes the determinant a simple product. The given condition, like x ≠ 3d, tells you which factor must vanish.

Answer from the NTA final answer key. Solution by PrepWiser.