Question

$ \Delta_{1}=\begin{vmatrix}x&b&b\\ a&x&b\\ a&a&x\end{vmatrix} and \Delta_{2} =\begin{vmatrix}x&b\\ a&x\end{vmatrix}$ are the given determinants, then

• A. $\Delta_{1}=3\left(\Delta_{2}\right)^{2}$
• B. $\frac{d}{d x}\left(\Delta_{1}\right)=3 \Delta_{2}$
• C. $\frac{d}{d x}\left(\Delta_{1}\right)=3\left(\Delta_{2}\right)^{2}$
• D. $\Delta_{1}=3 \Delta_{2}^{3 / 2}$

Answer 1

Answer: (B)

$ \Delta_{1} =x\left(x^{2}-a b\right)-b(a x-a b)+b\left(a^{2}-a x\right)$
$=x^{3}-3 a b x+a b^{2}+a^{2} b $
$ \therefore \frac{d}{d x}\left(\Delta_{1}\right)=3 x^{2}-3 a b=3\left(x^{2}-a b\right)=3 \Delta_{2} $