Question

If $\alpha ,\beta \ne 0$ and $f\left(n\right)={\alpha }^n+{\beta }^n$ and $\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|=K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$, then K is equal to

• A. $\alpha \beta$
• B. $\frac{1}{\alpha \beta }$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

Plan, Use the properly that, two determinants can be multiplied column - to - row or row - to - column, to write the given determinant as the product of two determinants and then expand.
Given, $f\left(n\right)={\alpha }^n+{\beta }^n,f\left(1\right)=\alpha +\beta ,f\left(2\right)={\alpha }^2+{\beta }^2,f\left(3\right)={\alpha }^3+{\beta }^3,f\left(4\right)={\alpha }^4+{\beta }^4$
Let $\Delta =\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|$
$\Rightarrow \Delta =\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
$=\left|\begin{array}{ccc}1.1+1.1+1.1& 1.1+1.\alpha +1.\beta & 1.1+1.{\alpha }^2+1.{\beta }^2\\ 1.1+1.\alpha +1.\beta & 1.1+\alpha .\alpha +\beta .\beta & 1.1+1.\alpha .{\alpha }^2+\beta .{\beta }^2\\ 1.1+1.{\alpha }^2+1.{\beta }^2& 1.1+\alpha .{\alpha }^2+\beta .{\beta }^2& 1.1+{\alpha }^2.{\alpha }^2+{\beta }^2.{\beta }^2\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$
On expanding, we get $\Delta =(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
But given, $\Delta =K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
Hence, $K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2=(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
$\therefore K=1$