Q. If $\mathrm{\Delta}_{1} = \left| \begin{matrix} x & b & b \\ a & x & b \\ a & a & x \end{matrix} \right|$ and $\mathrm{\Delta}_{2} = \left| \begin{matrix} x & b \\ a & x \end{matrix} \right|$ are the given determinants, then
A) $\mathrm{\Delta}_{1} = 3\left( \mathrm{\Delta}_{2} \right)^{2}$
B) $\frac{d}{dx}\left( \mathrm{\Delta}_{1} \right) = 3\mathrm{\Delta}_{2}$
C) $\frac{d}{dx}\left( \mathrm{\Delta}_{1} \right) = 2\mathrm{\Delta}_{2}$
D) $\mathrm{\Delta}_{1} = 3\mathrm{\Delta}_{2}^{3/2}$
Explanation
(b)
We have, $\mathrm{\Delta}_{1} = \left| \begin{matrix}
x & b & b \\
a & x & b \\
a & a & x
\end{matrix} \right| = x^{3} - 3abx + ab^{2} + a^{2}b$
$\Rightarrow \frac{d}{dx}\mathrm{\Delta}_{1} = 3\left( x^{2} - ab \right)\ and\ \mathrm{\Delta}_{2} = \left| \begin{matrix}
x & b \\
a & x
\end{matrix} \right| = x^{2} - ab$
$\therefore\frac{d}{dx}\left( \mathrm{\Delta}_{1} \right) = 3\left( x^{2} - ab \right) = 3\mathrm{\Delta}_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.