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(chatquestion)...10:30 AM ...0.0KB/s A GH ll 6D Determinants Q19: Choose the correct answer. 1 sin 0 1 Let A =-sin 0 1 sin 0 where 0s 0s 2n, -1 -sin e 1 then A. Det (A) = 0 B. Det (A) E (2, ∞) C. Det (A) E (2, 4) D. Det (A)E [2, 4] Solution : Answer: D sin e A= -sin 0 1 sin 0 -1 - sin e 1 :14| = 1(1+ sin 0)- sin 0(-sin 0 + sin 0) +1(sin 0 +1) =1+ sin 0+ sin0+1 %3D =2+2 sin? 0 = 2(1+ sin' 0) Now, 0 s 0 < 2N = -1 s sin 0 < 1 =0 s sin0< Isl+sin 0s2 =2s2(1+ sin 0)s 4 :Det (4) e[2,4] PAGE 143 The correct answer is D....(Ye kaise aaya underline wala anyone please help me)
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telegram-3359
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Jul 16, 2020
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