Question

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

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

Answer 1

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

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

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

$=(x^2-ab)+(x^2-ab)+(x^2-ab)=3(x^2-ab)$

and ${\Delta }_2=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|=x^2-ab$

Hence option B is correct