Question

If ${\Delta }_1=\left|\begin{array}{ccc}x& b& b\\ a& x& a\\ a& a& x\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|$ are the given determinats, then

• A. ${\Delta }_1=3{\left({\Delta }_2\right)}^2$
• B. $\frac{d}{dx}\left({\Delta }_1\right)=3{\Delta }_2$
• C. $\frac{d}{dx}\left({\Delta }_1\right)=3{\left({\Delta }_2\right)}^2$
• D. ${\Delta }_1=3{\left({\Delta }_2\right)}^{3/2}$

Answer 1

The correct option is B $\frac{d}{dx}\left({\Delta }_1\right)=3{\Delta }_2$

${\Delta }_1=\left|\begin{array}{ccc}x& b& b\\ a& x& a\\ a& a& x\end{array}\right|$

Differentiating w.r.t $x$, we get

$\frac{d}{dx}\left({\Delta }_1\right)=\left|\begin{array}{ccc}1& 0& 0\\ a& x& b\\ a& a& x\end{array}\right|+\left|\begin{array}{ccc}x& b& b\\ 0& 1& 0\\ a& a& x\end{array}\right|+\left|\begin{array}{ccc}x& b& b\\ a& x& a\\ 0& 0& 1\end{array}\right|$

$=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|+\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|+\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|=3{\Delta }_2$