Question

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3+a{x}^2+b=0$, $b\ne 0$ then the determinant $\Delta$, where
$\Delta =\left|\begin{array}{ccc}\frac{1}{\alpha }& \frac{1}{\beta }& \frac{1}{\gamma }\\ \frac{1}{\beta }& \frac{1}{\gamma }& \frac{1}{\alpha }\\ \frac{1}{\gamma }& \frac{1}{\alpha }& \frac{1}{\beta }\end{array}\right|$ equals

• A. $-b^3$
• B. $b^3-3c$
• C. $b^3+3c$
• D. $0$

Answer 1

Answer: (D)

$0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+a{x}^2+b=0$

Gives $\beta \gamma +\alpha \gamma +\alpha \beta =0$

$\Delta =\left|\begin{array}{ccc}\frac{1}{\alpha }& \frac{1}{\beta }& \frac{1}{\gamma }\\ \frac{1}{\beta }& \frac{1}{\gamma }& \frac{1}{\alpha }\\ \frac{1}{\gamma }& \frac{1}{\alpha }& \frac{1}{\beta }\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

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

Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$

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

$=-(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }){(\frac{1}{\gamma }-\frac{1}{\beta })}^2{(\frac{1}{\alpha }-\frac{1}{\beta })}^2=-\left(\frac{\beta \gamma +\alpha \gamma +\alpha \beta }{\alpha \beta \gamma }\right){\left(\frac{\beta -y}{\beta \gamma }\right)}^2{\left(\frac{\beta -\alpha }{\alpha \beta }\right)}^2=0$