Question

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

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

Answer 1

Answer: (B)

$\frac{d}{dx}{\Delta }_1=3{\Delta }_2$

Given

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

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

Expand along $R_1$

${\Delta }_1=x[x^2-ab]-b[ax-ab]+b[a^2-ax]$

$=x^3-abx-abx+a{b}^2+b{a}^2+b{a}^2-abx$

${\Delta }_1=x^3-3abx+ab(a+b)$

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

$=3x^2-3ab+0$

$\frac{d}{dx}\left({\Delta }_1\right)=3(x^2-ab)$ ....(1)

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

substituting in (1)

$\Rightarrow \frac{d}{dx}\left({\Delta }_1\right)=3\left({\Delta }_2\right)$

$\therefore \frac{d}{dx}\left({\Delta }_1\right)=3{\Delta }_2$