DeterminantHard
Question
Let A = 
such that |A| = 0, then maximum value of x + y + z is
such that |A| = 0, then maximum value of x + y + z isOptions
A.3
B.0
C.1
D.3
Solution
Let a = cos-1 x, b = cos-1 y, c = cos-1 z,
|A| = 0 ⇒
= (a + b + c) [1/2 {(a - b)2 + (b - c)2 + (c - a)2}] = 0
⇒ a + b + c = 0 ⇒ x = y = z = 1
|A| = 0 ⇒
= (a + b + c) [1/2 {(a - b)2 + (b - c)2 + (c - a)2}] = 0⇒ a + b + c = 0 ⇒ x = y = z = 1
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