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Q : 10        Evaluate  \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}

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We have determinant  \triangle = \begin{vmatrix} 1 & x &y \\ 1 &x+y &y \\ 1 &x &x+y \end{vmatrix}

Applying row transformations;  R_{1} \rightarrow R_{1}-R_{2}  and R_{2} \rightarrow R_{2}-R_{3} then  we have then;

\triangle = \begin{vmatrix} 0 & -y &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}

Taking out the common factor -y from the row first.

\triangle = -y\begin{vmatrix} 0 & 1 &0 \\ 0 &y &-x \\ 1 &x &x+y \end{vmatrix}

Expanding the remaining determinant;

-y[1(-x-o)] = xy

 

 

 

Posted by

Divya Prakash Singh

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