Question

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right|$ $=(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )(\alpha +\beta +\gamma )$

Answer 1

Consider, $\left|\begin{array}{ccc}\alpha & {\alpha }^2& \beta +\gamma \\ \beta & {\beta }^2& \gamma +\alpha \\ \gamma & {\gamma }^2& \alpha +\beta \end{array}\right|$

$C_3\to C_1+C_3$

$\left|\begin{array}{ccc}\alpha & {\alpha }^2& \alpha +\beta +\gamma \\ \beta & {\beta }^2& \alpha +\beta +\gamma \\ \gamma & {\gamma }^2& \alpha +\beta +\gamma \end{array}\right|$

Taking $\alpha +\beta +\gamma$ common from $C_3$

$=(\alpha +\beta +\gamma )\left|\begin{array}{ccc}\alpha & {\alpha }^2& 1\\ \beta & {\beta }^2& 1\\ \gamma & {\gamma }^2& 1\end{array}\right|$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$=(\alpha +\beta +\gamma )\left|\begin{array}{ccc}\alpha -\beta & {\alpha }^2-{\beta }^2& 0\\ \beta -\gamma & {\beta }^2-{\gamma }^2& 0\\ \gamma & {\gamma }^2& 1\end{array}\right|$

Taking $\alpha -\beta$ common from $R_1$ and $\beta -\gamma$ from $R_2$

$=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )\left|\begin{array}{ccc}1& \alpha +\beta & 0\\ 1& \beta +\gamma & 0\\ \gamma & {\gamma }^2& 1\end{array}\right|$

Expanding along the third column, we get

$=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )\left[1\right(\beta +\gamma -\alpha -\beta \left)\right]$

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