Mathematics - Determinant Question with Solution | TestHub

$f\left(1\right)=\alpha +\beta , f\left(2\right)={\alpha }^2+{\beta }^2, f\left(3\right)={\alpha }^3+{\beta }^3, f\left(4\right)={\alpha }^4+{\beta }^4$

So, the given determinant can be written as

$\left|\begin{array}{ccc}1+1+1& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$ $=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& \beta & {\beta }^2\end{array}\right|$

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

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

$={\left[\alpha \beta \left(\beta -\alpha \right)-\left(\beta -\alpha \right)\left(\beta +\alpha \right)+\left(\beta -\alpha \right)\right]}^2$

$={\left(\beta -a\right)}^2{\left[\mathit{\alpha \beta }-\beta -\alpha +1\right]}^2$

 $={\left(\beta -\alpha \right)}^2{\left(\alpha -1\right)}^2{\left(\beta -1\right)}^2$

Hence, $K=1$.