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

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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

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