DeterminantHard
Question
The least value of the product xyz for which the determinant
is non-negative is :
Options
A.−8
B.−1
C.−2√2
D.−16√2
Solution
⇒ xyz + 2 - y - x - z ≥ 0
⇒ xyz + 2 ≥ x + y + z ≥ 3 (xyz)1/3
put (xyz)1/3 = t
⇒ t3 - 3t + 2 ≥ 0
⇒ (t - 1)(t2 + t - 2) ≥ 0
⇒ (t - 1)2(t + 2) ≥ 0
⇒ t ≥ - 2
⇒ (xyz)1/3 ≥ - 2
⇒ xyz ≥ - 8
Create a free account to view solution
View Solution FreeMore Determinant Questions
Let A = be a matrix, then (det A) x (adj A-1) is equal to...If A = and B = , then...If a ≠ b, then the system of equations ax + by + bz = 0, bx + ay + bz = 0, bx + by + az = 0 will have a non-trivia...If A = (where bc ≠ 0) satisfies the equations x2 + k = 0, then...The system of linear equations$${x + y + z = 6 }{2x + 5y + az = 36 }{x + 2y + 3z = b }$$has...