Question

If
$\left|\begin{array}{ccc}(a-1{)}^2& a^2& (a+1{)}^2\\ (b-1{)}^2& b^2& (b+1{)}^2\\ (c-1{)}^2& c^2& (c+1{)}^2\end{array}\right|=k\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|,$
then the value of $k$ is

• A. $4$
• B. $-4$
• C. $1$
• D. $-2$

Answer 1

The correct option is B $-4$

$C_3\to C_3-C_1$
$\Delta =\left|\begin{array}{ccc}(a-1{)}^2& a^2& 4a\\ (b-1{)}^2& b^2& 4b\\ (c-1{)}^2& c^2& 4c\end{array}\right|=4\left|\begin{array}{ccc}(a-1{)}^2& a^2& a\\ (b-1{)}^2& b^2& b\\ (c-1{)}^2& c^2& c\end{array}\right|$

$C_1\to C_1-C_2+2C_3$
$\Delta =4\left|\begin{array}{ccc}1& a^2& a\\ 1& b^2& b\\ 1& c^2& c\end{array}\right|=-4\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
$\Rightarrow k=-4$