Question
Download Solution PDFयदि रैखिक समीकरणों का निकाय
2x + 2y + 3z = a
3x - y + 5z = b
x - 3y + 2z = c
जहाँ a, b, c शून्येतर वास्तविक संख्याएँ हैं, के एक से अधिक हल हैं, तो:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFयदि समीकरणों के निकाय के एक से अधिक हल हैं, तो D = 0 और D1 = D2 = D3 = 0 है।
दिए गए समीकरणों के समुच्चय से, हम बनाते हैं,
D1 = 0
\(\Rightarrow \left| \begin{matrix} a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2 \\ \end{matrix} \right|=0\)
अब,
\(⇒ a(-2 + 15) - 2(2b - 5c) + 3(-3b + c) = 0\)
\(⇒ 13a - 4b + 10c - 9b + 3c = 0 \)
\(⇒ 13a - 13b + 13c = 0 \)
\(⇒ a - b + c = 0 \)
\(∴ b - a - c = 0\)Last updated on Jul 11, 2025
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