Question

If $y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$, then $\frac{dy}{dx}$ is equal to

• A. $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$
• B. $\left|\begin{array}{ccc}l& m& n\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ a& b& c\end{array}\right|$
• C. $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& l& a\\ g^{\prime }\left(x\right)& m& b\\ h^{\prime }\left(x\right)& n& c\end{array}\right|$
• D. $\left|\begin{array}{ccc}l& m& n\\ a& b& c\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\end{array}\right|$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$
C $\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& l& a\\ g^{\prime }\left(x\right)& m& b\\ h^{\prime }\left(x\right)& n& c\end{array}\right|$
D $\left|\begin{array}{ccc}l& m& n\\ a& b& c\\ f^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\end{array}\right|$

$y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$
$\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{{}^{\prime }}\left(x\right)& g^{{}^{\prime }}\left(x\right)& h^{{}^{\prime }}\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ 0& 0& 0\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ 0& 0& 0\end{array}\right|$
$\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{{}^{\prime }}\left(x\right)& g^{{}^{\prime }}\left(x\right)& h^{{}^{\prime }}\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|\Rightarrow$ option(A)
$R_1\leftrightarrow R_2$ and then$R_2\leftrightarrow R_3$
$\frac{dy}{dx}=\left|\begin{array}{ccc}l& m& n\\ a& b& c\\ f^{{}^{\prime }}\left(x\right)& g^{{}^{\prime }}\left(x\right)& h^{{}^{\prime }}\left(x\right)\end{array}\right|\Rightarrow$option (D)
Now, $\left|A\right|=\left|A^T\right|$
$\therefore \frac{dy}{dx}={\left|\begin{array}{ccc}{f}^{{}^{\prime }}\left(x\right)& g^{{}^{\prime }}\left(x\right)& h^{{}^{\prime }}\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|}^T$
$=\left|\begin{array}{ccc}{f}^{{}^{\prime }}\left(x\right)& l& a\\ g^{{}^{\prime }}\left(x\right)& m& b\\ h^{{}^{\prime }}\left(x\right)& n& c\end{array}\right|\Rightarrow$option (C)
Option(B) is not correct
$\because$ it will also contain negative sign
First and Second rows are interchanged.