Question

if f(x)= $\left|\begin{array}{ccc}3& 1& 2\\ a\left(x\right)& b\left(x\right)& c\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|\text{,then f'(x)=}\left|\begin{array}{ccc}3& 1& 2\\ a^{\prime }\left(x\right)& b^{\prime }\left(x\right)& c^{\prime }\left(x\right)\\ g^{\prime }\left(x\right)& h^{\prime }\left(x\right)& k^{\prime }\left(x\right)\end{array}\right|$

• A. True
• B. False

Answer 1

The correct option is B

False

we know if a matrix contains elements as functions of the derivative of the determinant will be the sum of determinants with each determinant having only one of its rows differentiated. So in this case,

$F^{\prime }\left(x\right)=\left|\begin{array}{ccc}3& 1& 2\\ a\left(x\right)& b\left(x\right)& c\left(x\right)\\ g^{\prime }\left(x\right)& h^{\prime }\left(x\right)& k^{\prime }\left(x\right)\end{array}\right|+\left|\begin{array}{ccc}3& 1& 2\\ a^{\prime }\left(x\right)& b^{\prime }\left(x\right)& c^{\prime }\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|+\left|\begin{array}{ccc}0& 0& 0\\ a\left(x\right)& b\left(x\right)& c\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|$

$=\left|\begin{array}{ccc}3& 1& 2\\ a\left(x\right)& b\left(x\right)& c\left(x\right)\\ g^{\prime }\left(x\right)& h^{\prime }\left(x\right)& k^{\prime }\left(x\right)\end{array}\right|+\left|\begin{array}{ccc}3& 1& 2\\ a^{\prime }\left(x\right)& b^{\prime }\left(x\right)& c^{\prime }\left(x\right)\\ g\left(x\right)& h\left(x\right)& k\left(x\right)\end{array}\right|$

Therefore the given option is false.