Question

Simplify and fine the value -

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

Answer 1

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

$R_1\to R_1-R_2$

$R_2\to R_2-R_3$

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

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

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

$R_1\to R_1-R_2$

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

$=0$