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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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## QuestionMathsClass 12

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

$${R_1} \to {R_1} - {R_3}$$$${R_2} \to {R_2} - {R_3}$$
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
$$= (\alpha - \gamma )(\beta - \gamma )\left[ { - (\alpha - \beta )(\alpha + \beta + \gamma )} \right]$$
$$= (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )$$