Question

Suppose $\alpha ,\beta$ are two real numbers and $f(n)={\alpha }^n+{\beta }^n$.
Let $\Delta =\left|\begin{array}{ccc}3& 1+f(1)& 1+f(2)\\ 1+f(1)& 1+f(2)& 1+f(3)\\ 1+f(2)& 1+f(3)& 1+f(4)\end{array}\right|$
If $\Delta =k(\alpha -1{)}^2(\beta -1{)}^2(\alpha -\beta {)}^2$, then $k$ is equal to

• A. 1
• B. $4\alpha \beta$
• C. 9
• D. ${\alpha }^2{\beta }^2$

Answer 1

Answer: (A)

$\Delta =\left|\begin{array}{ccc}1+1+1& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|={\Delta }_1^2$
where${\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
Applying $C_3\to C_3-C_2$ and $C_2\to C_2-C_1$, we get
${\Delta }_1=\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -\alpha \\ 1& {\alpha }^2-1& {\beta }^2-{\alpha }^2\end{array}\right|$
$=(\alpha -1)(\beta -\alpha )\left|\begin{array}{cc}1& 1\\ \alpha +1& \beta +\alpha \end{array}\right|$
$=(\alpha -1)(\beta -1)(\beta -\alpha )$
$\text{ Thus, }\Delta =(\alpha -1{)}^2(\beta -1{)}^2(\alpha -\beta {)}^2$
$\therefore k=1$