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  • Using properties of determinants, prove that det alpha alpha^2 beta+gamma beta beta^2 gamma+alpha gamma gamma^2 alpha+beta = (beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Q : 11        Using properties  of determinants, prove that

                  \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}=(\beta -\gamma )(\gamma -\alpha )(\alpha -\beta )(\alpha +\beta +\gamma )

Answers (1)

best_answer

Given determinant  \triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta & \beta ^2 &\gamma +\alpha \\ \gamma &\gamma ^2 &\alpha +\beta \end{vmatrix}

Applying Row transformations; and  R_{3} \rightarrow R_{3}-R_{1}, then we have;

\triangle = \begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ \beta -\alpha & \beta ^2 - \alpha^2 &\alpha - \beta \\ \gamma-\alpha &\gamma ^2-\alpha^2 &\alpha -\gamma \end{vmatrix}

= (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha ^2 &\beta +\gamma \\ 1 & \beta + \alpha &-1 \\ 0 &\gamma-\beta &0\end{vmatrix}

Expanding the remaining determinant;

= (\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]

= (\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)

=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)

hence the given result is proved.

Posted by

Divya Prakash Singh

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