सारणिकों के गुणधर्मों का प्रयोग करके सिद्ध कीजिए :

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

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$\Delta=\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

$=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & x+4 & 0 \\ 2 x & 0 & x+4\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow C_{2}-C_{1}, \mathrm{C}_{3} \rightarrow C_{3}-C_{1},$ we have

$\Delta=(5 x+4)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & -x+4 & 0 \\ 2 x & 0 & -x+4\end{array}\right|$

$=(5 x+4)(4-x)(4-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1\end{array}\right|$

Expanding along $C_{3},$ we have:

$\Delta=(5 x+4)(4-x)^{2}\left|\begin{array}{cc}1 & 0 \\ 2 x & 1\end{array}\right|$

$=(5 x+4)(4-x)^{2}$

Hence, the given result is proved.

Similar Questions

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यदि  $a, b$ और  $ c$   तीन अशून्य वास्तविक संख्यायें हैं, तो $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2}{c^2}}&{bc}&{b + c}\\{{c^2}{a^2}}&{ca}&{c + a}\\{{a^2}{b^2}}&{ab}&{a + b}\end{array}\,} \right| $ =

यदि ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, तो $\sum\limits_{r = 1}^n {{D_r}} $ का मान है

सिद्ध कीजिए कि सारणिक

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$