Misc 19 - Chapter 4 Class 12 Determinants

Transcript

Misc. 19 Choose the correct answer. Let A = 1 sin 1 sin 1 sin 1 sin 1 , where 0 2 , then A. Det (A) = 0 B. Det (A) (2, C. Det (A) (2, 4) D. Det (A) [2, 4] A = 1 sin 1 sin 1 sin 1 sin 1 |A| = 1 sin 1 sin 1 sin 1 sin 1 = 1 1 sin sin 1 sin sin sin 1 1 + 1 sin 1 1 sin = 1 (1 + sin2 ) sin ( sin + sin ) + 1 (sin2 + 1) = 1 (1 + sin2 ) sin (0) + 1(sin2 + 1) = 2 (1 + sin2 ) Thus, |A| = 2 (1 + sin2 ) We know that 1 sin 1 So, value of sin can be from 1 to 1 Suppose, Hence, value of sin2 can be from 0 to 1 (negative not possible) Thus 2 |A| 4 |A| [2 , 4] Det (A) [2 , 4] Thus D is the correct answer