Last updated at Jan. 23, 2020 by Teachoo

Transcript

Misc 19 Choose the correct answer. Let A = [■8(1&sinθ&1@−sinθ&1&sinθ@−1&〖−sin〗θ&1)] , where 0 ≤ θ≤ 2π, then A. Det (A) = 0 B. Det (A) ∈ (2, ∞) C. Det (A) ∈ (2, 4) D. Det (A)∈ [2, 4] A = [■8(1&sinθ&1@−sinθ&1&sinθ@−1&〖−sin〗θ&1)] |A| = |■8(1&sinθ&1@−sinθ&1&sinθ@−1&〖−sin〗θ&1)| = 1 |■8(1&sinθ@−sinθ&1)| – sin θ |■8(−sinθ&sinθ@−1&1)| + 1 |■8(−sinθ&1@−1&〖−sin〗θ )| = 1 (1 + sin2 θ) – sin θ (–sin θ + sin θ) + 1 (sin2 θ + 1) = (1 + sin2 θ) –sin θ × 0 + (1 + sin2 θ) = 2 (1 + sin2 θ) Thus, |A| = 2 (1 + sin2 θ) We know that –1 ≤ sin θ ≤ 1 So, value of sin θ can be from –1 to 1 Suppose, Hence, value of sin2 θ 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 sin θ = –1 sin2 θ = 1 sin θ = 1 sin2 θ = 1 sin θ = 0 sin2 θ = 0 Putting sin2 θ = 0 in |A| |A| = 2(1 + 0) |A| = 2 Thus minimum value of |A| is 2 Putting sin2 θ = 1 in |A| |A| = 2 (1 + 1) = 2 (2) = 4 Thus maximum value of |A| is 4

Miscellaneous

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Misc. 2 Important Deleted for CBSE Board 2021 Exams only

Misc 3

Misc 4

Misc 5

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 Important Deleted for CBSE Board 2021 Exams only

Misc 12 Important Deleted for CBSE Board 2021 Exams only

Misc. 13 Deleted for CBSE Board 2021 Exams only

Misc 14 Deleted for CBSE Board 2021 Exams only

Misc. 15 Important Deleted for CBSE Board 2021 Exams only

Misc. 16 Important

Misc 17 Important Deleted for CBSE Board 2021 Exams only

Misc 18

Misc 19 Important You are here

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.