यदि left| {begin{array}{*{20}{c}}{x - 4} | vedclass

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Question

(2) Here, $\left| {\begin{array}{*{20}{c}}
{x - 4}&{2x}&{2x}\
{2x}&{x - 4}&{2x}\
{2x}&{2x}&{x - 4}
\end{array}} i

Vedclass · Accepted Answer

$\left( { - 4,5} \right)$

Vedclass · Answer

(2) Here, $\left| {\begin{array}{*{20}{c}}
{x - 4}&{2x}&{2x}\
{2x}&{x - 4}&{2x}\
{2x}&{2x}&{x - 4}
\end{array}} ight| = \left( {A + Bx} ight){\left( {x - A} ight)^2}$

Put $x = 0 \Rightarrow \left| {\begin{array}{*{20}{c}}
{ - 4}&0&0\
0&{ - 4}&0\
0&0&{ - 4}
\end{array}} ight| = {A^3} \Rightarrow {A^3} = {\left( { - 4} ight)^3}$

$ \Rightarrow A =  - 4$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
{x - 4}&{2x}&{2x}\
{2x}&{x - 4}&{2x}\
{2x}&{2x}&{x - 4}
\end{array}} ight| = \left( {Bx - 4} ight){\left( {x - 4} ight)^2}$

Now take $x$ common from both the sides

$	herefore \left| {\begin{array}{*{20}{c}}
{1 - \frac{4}{x}}&{2x}&{2x}\
{2x}&{1 - \frac{4}{x}}&{1 - \frac{4}{x}}\
{2x}&{2x}&{2x}
\end{array}} ight| = \left( {B - \frac{4}{x}} ight){\left( {1 + \frac{4}{x}} ight)^2}$

यदि $\left| {\begin{array}{*{20}{c}}{x - 4}&{2x}&{2x}\\{2x}&{x - 4}&{2x}\\{2x}&{2x}&{x - 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},$ तो क्रमित युग्म $(A, B)$ बराबर है

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