Question
Download Solution PDFयदि \(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\) है, तो निम्नलिखित में से कौन-सा सही है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
यदि \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) है, तो A की सारणिक निम्न दी गयी है:
|A| = a11 × {(a22 × a33) – (a23 × a32)} - a12 × {(a21 × a33) – (a23 × a31)} + a13 × {(a21 × a32) – (a22 × a31)}
गणना:
दिया गया है:
\(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\)
R1 के साथ विस्तृत करने पर, हमें निम्न प्राप्त होता है
⇒ x (x2 – 0) – y (0 – y2) + 0 = 0
⇒ x3 + y3 = 0
\(\Rightarrow {\rm{\;}}{{\rm{y}}^3}{\rm{\;}}\left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right){\rm{\;}} = {\rm{\;}}0{\rm{\;}}\)
\(\therefore \left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right) = {\rm{\;}}0{\rm{\;and\;}}{{\rm{y}}^3} \ne 0\)
\(\Rightarrow \frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} = {\rm{}} - 1\)
\(\Rightarrow \frac{{\rm{x}}}{{\rm{y}}} = {\left( { - 1} \right)^{\frac{1}{3}}}\)
अतः x/y, -1 का एक घनमूल है।
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