Question

Let $\alpha ,\beta$ be the roots of the equation $a{x}^2+bx+c=0$. Let $S_n={\alpha }^n+{\beta }^n$ for $n\ge 1$ Evaluate the determinant $\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|$

• A. $\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$
• B. $\frac{(a+b+c{)}^2(b^2-4ac)}{a^2}$
• C. $\frac{(a+b+c{)}^2(b^2+4ac)}{a^2}$
• D. $\frac{(a+b+c{)}^2(b^2+4ac)}{a^4}$

Answer 1

The correct option is A $\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$

Since $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$
$\therefore \alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$
Let
$\Delta =\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|$
$=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^3& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^3\end{array}\right|$
$=({\Delta }_1\times {\Delta }_1)$ (say)
$\therefore \Delta ={\Delta }_1^2$ (i)
$\therefore {\Delta }_1=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, then
${\Delta }_1=\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -1\\ 1& {\alpha }^2-1& {\beta }^2-1\end{array}\right|$
Expanding along $R_1$
$\therefore {\Delta }_1=\left|\begin{array}{cc}\alpha -1& \beta -1\\ {\alpha }^2-1& {\beta }^2-1\end{array}\right|$
$=(\alpha -1)(\beta -1)\left|\begin{array}{cc}1& 1\\ \alpha +1& \beta +1\end{array}\right|$
$=\{\alpha \beta -(\alpha +\beta )+1\}(\beta -\alpha )$
$=\{\alpha \beta -(\alpha +\beta )+1\}\sqrt{\left\{\right(\alpha +\beta {)}^2-4\alpha \beta \}}$
$=(\frac{c}{a}+\frac{b}{a}+1)\sqrt{(\frac{b^2}{a^2}-\frac{4c}{a})}$
${\Delta }_1=\frac{(a+b+c)\sqrt{(b^2-4ac)}}{a^2}$
$\therefore {\Delta }_1^2=\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$
Hence, $\Delta ={\Delta }_1^2=\frac{(a+b+c{)}^2(b^2-4ac)}{a^4}$