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Using properties of determinant prove that . \(\left| {\begin{array}{*{20}{c}} \alpha &{{\alpha ^2}}&{\beta + \gamma } \\ \beta &{{\beta ^2}}&{\gamma + \alpha } \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right| = (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )\)
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Using properties of determinant prove that . \(\left| {\begin{array}{*{20}{c}} \alpha &{{\alpha ^2}}&{\beta + \gamma } \\ \beta &{{\beta ^2}}&{\gamma + \alpha } \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right| = (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )\)

Answer

\({R_1} \to {R_1} - {R_3}\)\({R_2} \to {R_2} - {R_3}\)
L.H.S \( = \left| {\begin{array}{*{20}{c}} {\alpha - \gamma }&{{\alpha ^2} - {\gamma ^2}}&{\beta + \gamma - \alpha - \beta } \\ {\beta -\gamma }&{{\beta ^2} - {\gamma ^2}}&{\gamma + \alpha - \alpha - \beta } \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right|\)
\( = \left| {\begin{array}{*{20}{c}} {\alpha - \gamma }&{(\alpha + \gamma )}(\alpha-\gamma)&{(\gamma - \alpha )} \\ {\beta - \gamma }&{(\beta - \gamma )(\beta + \gamma )}&{\gamma - \beta } \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right|\)
\( = (\alpha - \gamma )(\beta - \gamma )\left| {\begin{array}{*{20}{c}} 1&{\alpha + \gamma }&{ - 1} \\ 1&{\beta + \gamma }&-1 \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right|\) \([Taking\ (\alpha-\gamma)\ common \ from R_1\ and(\beta-\gamma)\ common \ from\ R_2]\)
\({R_1} \to {R_1} - {R_2}\)
\( = (\alpha - \gamma )(\beta - \gamma )\left| {\begin{array}{*{20}{c}} 0&{\alpha - \beta }&0 \\ 1&{\beta + \gamma }&{ - 1} \\ \gamma &{{\gamma ^2}}&{\alpha + \beta } \end{array}} \right|\)
Expending along R1
\( = (\alpha - \gamma )(\beta - \gamma )\left[ { - (\alpha - \beta )(\alpha + \beta + \gamma )} \right]\)
\( = (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )\)
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