Question

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

• A. ${\Delta }_1=3{\left({\Delta }_2\right)}^2$
• 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. ${\Delta }_1=3{\Delta }_2^{3/2}$

Answer 1

Answer: (B)

${\Delta }_1=x\left(x^2-ab\right)-b(ax-ab)+b\left(a^2-ax\right)$
$=x^3-3abx+a{b}^2+a^{2}b$
$\therefore \frac{d}{dx}\left({\Delta }_1\right)=3x^2-3ab=3\left(x^2-ab\right)=3{\Delta }_2$