बिना प्रसरण किए और सारणिकों के गुणधर्मो का प्रयोग करके सिद्ध कीजिए।

$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

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$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=\left|\begin{array}{lll}2 & 7 & 63+2 \\ 3 & 8 & 72+3 \\ 5 & 9 & 81+5\end{array}\right|$

$=\left|\begin{array}{lll}2 & 7 & 63 \\ 3 & 8 & 72 \\ 5 & 9 & 81\end{array}\right|+\left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$

$=\left|\begin{array}{lll}2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9)\end{array}\right|+0$        [ Two Coloumms are identical]

$=9\left|\begin{array}{lll}2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9\end{array}\right|$

$=0$         [ Two Coloumns are identical]

Similar Questions

यदि ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ , तब $\sum\limits_{n = 1}^N {{U_n}} $ का मान है

सारणिकों के गुणधर्मो का प्रयोग करके निम्नलिखित प्रश्न को सिद्ध कीजिए :

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x)$

यदि $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}\,} \right| = k\,abc{(a + b + c)^3}$, तो $k$ का मान है

$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $

  • [IIT 1980]

यदि $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$,  तब $k = $