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

Answer 1

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

${\Delta }_1=\left|\begin{array}{ccc}x& b& b\\ a& x& b\\ a& a& x\end{array}\right|=x(x^2-ab)-b(ax-ab)+b(a^2-ax)$
$\Rightarrow {\Delta }_1=x^3-3abx+a{b}^2+a^{2}b$
$\Rightarrow \frac{d}{dx}{\Delta }_1=3(x^2-ab)...\left(1\right)$
Now, ${\Delta }_2=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|=x^2-ab...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$\frac{d}{dx}{\Delta }_1=3{\Delta }_2$