Question
Download Solution PDFजर ω हे एकतेचे घनमूळ असेल, तर समीकरणाचे मूळ काय आहे?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
जर ω हे एकतेचे घनमूळ असेल, म्हणजेच ω3 = 1.
तर 1 + ω + ω2 = 0.
ω4 = ω3ω = ω [∵ ω3 =1]
गणना:
दिलेले आहे:
C'1 = C1 + C2
R'2 = R2 - R1 आणि R'3 = R3 - R1
पहिल्या स्तंभासह विस्तारत आहे:
∴ x[(x + ω2 - ω)(x + ω - ω2) - (1 - ω)(1 - ω2)]
∴ x[x2 + ωx - ω2x + ω2x + ω3 - ω4 - ωx - ω2 + ω3 - 1 + ω2 + ω - ω3]
∴ x3 = 0 (∵ ω3 = 1 आणि ω4 = ω3ω ⇒ ω)
∴ x = 0.
Last updated on May 9, 2025
-> The BPSC Assistant Professor last date to apply online has been extended to 15th May 2025 (Advt. No. 04/2025 to 28/2025).
-> The BPSC Assistant Professor Notification 2025 has been released for 1711 vacancies under Speciality (Advt. No.04/2025 to 28/2025).
-> The recruitment is ongoing for 220 vacancies (Advt. No. 01/2024 to 17/2024).
-> The salary under BPSC Assistant Professor Recruitment is approximately Rs 15600-39100 as per Pay Matrix Level 11.
-> The selection is based on the evaluation of academic qualifications & work experience and interview.
-> Prepare for the exam using the BPSC Assistant Professor Previous Year Papers. For mock tests attempt the BPSC Assistant Professor Test Series.