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

Step-by-step solution

Given x ,   y ,   z are in arithmetic progression, hence 2 y = x + z , also 3 4 2 x 4 5 2 y 5 k z = 0 Applying R 2 → R 1 + R 3 - 2 R 2 ⇒ 3 4 2 x 0 k - 6 2 0 5 k z = 0 ⇒ ( k - 6 2 ) ( 3 z - 5 x ) = 0 ⇒ ( k - 6 2 ) = 0 or ( 3 z - 5 x ) = 0 If 3 z - 5 x = 0 ⇒ 3 ( x + 2 d ) - 5 x = 0 ⇒ x = 3 d (Not possible) ⇒ k = 6 2 ⇒ k 2 = 72 .