Question
Download Solution PDFনিম্নলিখিত নির্ণায়কটির উৎপাদক বিশ্লেষিত রূপ হল:
\(\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 1&m&{{m^2}}\\ 1&n&{{n^2}} \end{array}} \right|\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF\(\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 1&m&{{m^2}}\\ 1&n&{{n^2}} \end{array}} \right|\)
R2 → R2 - R1 প্রয়োগ করে পাই
\(\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 0&{m - l}&{{m^2} - {l^2}}\\ 1&n&{{n^2}} \end{array}} \right|\)
R3 → R3 - R1 প্রয়োগ করে পাই
\(\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 0&{m - l}&{{m^2} - {l^2}}\\ 0&{n - l}&{{n^2} - {l^2}} \end{array}} \right|\)
\(\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 0&{m - l}&{\left( {m - l} \right)\left( {m\; + \;l} \right)}\\ 0&{n - l}&{\left( {n - l} \right)\left( {n\; + \;l} \right)} \end{array}} \right|\)
\(\left( {m - l} \right)\left( {n - l} \right)\left| {\begin{array}{*{20}{c}} 1&l&{{l^2}}\\ 0&1&{\left( {m\; + \;l} \right)}\\ 0&1&{\left( {n\; + \;l} \right)} \end{array}} \right|\)
এখন, a11 থেকে বিস্তার করে পাই।
\(\left( {m - l} \right)\left( {n - l} \right).1.\left[ {\begin{array}{*{20}{c}} 1&{m\; + \;l}\\ 1&{n\; + \;l} \end{array}} \right]\)
= (m - l)(n - l)(n + l - m - l)
= (m - l)(n - l)(n - m)
Last updated on Jul 7, 2025
->UPSC NDA Application Correction Window is open from 7th July to 9th July 2025.
->UPSC had extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.