Using properties of deteminants, find the value of k if  \begin{vmatrix} x & y & x+y\\ y& x+y &x \\ x+y & x & y \end{vmatrix}= k\left ( x^{3}+y^{3} \right )

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} x & y &x+y \\ y& x+y &x \\ x+y&x &y \end{vmatrix}= k\left [ x^{3}+y^{3} \right ]
By R_{1}\rightarrow R_{1}+R_{2}+R_{3}
\begin{vmatrix} 2\left ( x+y \right ) & 2\left ( x+y \right ) &2\left ( x+y \right ) \\ y& x+y &x \\ x+y&x &y \end{vmatrix}= k\left [ x^{3}+y^{3} \right ]
By c_{2}\rightarrow c_{2}-c_{1}\; \S \; c_{3}\rightarrow c_{3}-c_{1}
\begin{vmatrix} 2\left ( x+y \right ) & 0 &0\\ y& x &x-y \\ x+y&-y &-x \end{vmatrix}= k\left [ x^{3}+y^{3} \right ]
Expanding along R1
2\left ( x+y \right )\left \{ -x^{2}+xy-y^{2} \right \}= k\left ( x^{3}+y^{3} \right )
\Rightarrow -2\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )= k\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )
on comparing both sides we get  k= -2



 

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