Question

lf $D_1$ $=\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|$ & $D_2=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ then:

• A. $D_1=D_2^2$
• B. $D_1^2=D_2$
• C. $D_1=D_2$
• D. $D_1=2D_2$

Answer 1

The correct option is C $D_1=D_2$

Given,$D_1=\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|$ and $D_2=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
$D_1=\frac{1}{abc}\left|\begin{array}{ccc}a& a^2& abc\\ b& b^2& abc\\ c& c^2& abc\end{array}\right|$
take abc common from $3^{rd}$ column
$D_1=\frac{abc}{abc}\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|=\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$
then even times swap columns of $D_1$,
$C_3\leftrightarrow C_2$ and then $C_1\leftrightarrow C_2$
So,$D_1=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=D_2.$
$\Rightarrow D_1=D_2$