Misc 14

Last updated at Dec. 8, 2016 by Teachoo

Misc 14 - Using properties of determinants, prove - Class 12 - Solving by simplifying det.

Transcript

Misc 14 Using properties of determinants, prove that: 1﷮1+p﷮1+p+q﷮2﷮3+2p﷮4+3p+2q﷮3﷮6+3p﷮10+6p+3q﷯﷯ = 1 Taking L.H.S 1﷮1+p﷮1+p+q﷮2﷮3+2p﷮4+3p+2q﷮3﷮6+3p﷮10+6p+3q﷯﷯ Applying R2→ R2 – 2R1 = 1﷮1+p﷮1+p+q﷮𝟐−𝟐(𝟏)﷮3+2p−2(1+p)﷮4+3𝑝+2𝑝−2(1+𝑝+𝑞)﷮3﷮6+3p−2﷮10+6p+3q﷯﷯ = 1﷮1+p﷮1+p+q﷮𝟎﷮1﷮2+𝑝﷮3﷮6+3p﷮10+6p+3q﷯﷯ Applying R3→ R3 – 3R2 = 1﷮1+p﷮1+p+q﷮0﷮1﷮2+p﷮𝟑−𝟑(𝟏)﷮6+3𝑝−3(1+𝑝)﷮10+6𝑝+3𝑞−3(1+𝑝+𝑞)﷯﷯ = 1﷮1+p﷮1+p+q﷮0﷮1﷮2+p﷮𝟎﷮3﷮7+3𝑝﷯﷯ Expanding determinant along C1 = 1 1﷮2+p﷮3﷮7+3p﷯﷯ – 0 1+𝑝﷮2+p﷮3﷮7+3p﷯﷯ + 0 1+𝑝﷮1+p+q﷮1﷮2+p﷯﷯ = 1 1﷮2+p﷮3﷮7+3p﷯﷯ – 0 + 0 = 1 (7 + 3p – 3 (2 + p)) = 7 + 3p – 6 – 3p = 1 = R.H.S Hence Proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.