Misc 5 - Solve |x+a x x x x+a x x x x+a| = 0 - Class 12 - Miscellaneous

Misc 5 - Chapter 4 Class 12 Determinants - Part 2
Misc 5 - Chapter 4 Class 12 Determinants - Part 3

Remove Ads Share on WhatsApp

Transcript

Question 3 Solve the equations š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ = 0, a ≠ 0 Solving āˆ† = š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ Applying R1 → R1 + R2 + R3 = š‘„+š‘Ž+š‘„+š‘„ļ·®š‘„+š‘„+š‘Ž+š‘„ļ·®š‘„+š‘„+š‘„+š‘Žļ·®š‘„ļ·®š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ = šŸ‘š’™+š’‚ļ·®šŸ‘š’™+š’‚ļ·®šŸ‘š’™+š’‚ļ·®š‘„ļ·®š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ Taking (3x + a) common from R1 = (3x + a) 1ļ·®1ļ·®1ļ·®š‘„ļ·®š‘„+š‘Žļ·®š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ Applying C2 → C2 – C1 = (3x + a) 1ļ·®šŸāˆ’šŸļ·®1ļ·®š‘„ļ·®š‘„+š‘Žāˆ’š‘„ļ·®š‘„ļ·®š‘„ļ·®š‘„āˆ’š‘„ļ·®š‘„+š‘Žļ·Æļ·Æ = (3x+ a) 1ļ·®šŸŽļ·®1ļ·®š‘„ļ·®š‘Žļ·®š‘„ļ·®š‘„ļ·®0ļ·®š‘„+š‘Žļ·Æļ·Æ Applying C3 →C3 – C1 = (3x + a) 1ļ·®0ļ·®šŸāˆ’šŸļ·®š‘„ļ·®š‘Žļ·®š‘„āˆ’š‘„ļ·®š‘„ļ·®0ļ·®š‘„+š‘Žāˆ’š‘„ļ·Æļ·Æ = (3x + a) 1ļ·®0ļ·®šŸŽļ·®š‘„ļ·®š‘Žļ·®0ļ·®š‘„ļ·®0ļ·®š‘Žļ·Æļ·Æ = (3x+ a) 1 š‘Žļ·®0ļ·®0ļ·®š‘Žļ·Æļ·Æāˆ’0 š‘„ļ·®0ļ·®š‘„ļ·®0ļ·Æļ·Æ+0 š‘„ļ·®š‘Žļ·®š‘„ļ·®0ļ·Æļ·Æļ·Æ = (3x+ a) 1 š‘Žļ·®0ļ·®0ļ·®š‘Žļ·Æļ·Æāˆ’0+0ļ·Æ = (3x+ a) (1(a2 – 0) – 0 + 0) = (3x+ a) (a2) ∓ āˆ† = (3x+ a) (a2) Given āˆ† = 0 ∓ (3x + a) a2 = 0

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo