

Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
Misc 14 Using properties of determinants, prove that: 11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q = 1 Taking L.H.S 11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q Applying R2→ R2 – 2R1 = 11+p1+p+q𝟐−𝟐(𝟏)3+2p−2(1+p)4+3𝑝+2𝑝−2(1+𝑝+𝑞)36+3p−210+6p+3q = 11+p1+p+q𝟎12+𝑝36+3p10+6p+3q Applying R3→ R3 – 3R2 = 11+p1+p+q012+p𝟑−𝟑(𝟏)6+3𝑝−3(1+𝑝)10+6𝑝+3𝑞−3(1+𝑝+𝑞) = 11+p1+p+q012+p𝟎37+3𝑝 Expanding determinant along C1 = 1 12+p37+3p – 0 1+𝑝2+p37+3p + 0 1+𝑝1+p+q12+p = 1 12+p37+3p – 0 + 0 = 1 (7 + 3p – 3 (2 + p)) = 7 + 3p – 6 – 3p = 1 = R.H.S Hence Proved
Miscellaneous
Misc. 2 Important Not in Syllabus - CBSE Exams 2021
Misc 3
Misc 4
Misc 5
Misc 6 Important
Misc 7 Important
Misc 8
Misc 9
Misc 10
Misc 11 Important Not in Syllabus - CBSE Exams 2021
Misc 12 Important Not in Syllabus - CBSE Exams 2021
Misc. 13 Not in Syllabus - CBSE Exams 2021
Misc 14 Not in Syllabus - CBSE Exams 2021 You are here
Misc. 15 Important Not in Syllabus - CBSE Exams 2021
Misc. 16 Important
Misc 17 Important Not in Syllabus - CBSE Exams 2021
Misc 18
Misc 19 Important
Matrices and Determinants - Formula Sheet and Summary
About the Author