Question

If $D=\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$ and $D^{\prime }=\left|\begin{array}{ccc}b+c& c+a& a+b\\ a+b& b+c& c+a\\ c+a& a+b& b+c\end{array}\right|$ then $D^{\prime }=$

• A. $2D$
• B. $D$
  
• C. $3D$
• D. $6D$

Answer 1

Answer: (A)

$2D$

$D^{\prime }=\left|\begin{array}{ccc}b+c& c+a& a+b\\ a+b& b+c& c+a\\ c+a& a+b& b+c\end{array}\right|$
$D^{\prime }=\left|\begin{array}{ccc}b& c+a& a+b\\ a& b+c& c+a\\ c& a+b& b+c\end{array}\right|+\left|\begin{array}{ccc}c& c+a& a+b\\ b& b+c& c+a\\ a& a+b& b+c\end{array}\right|$
$D^{\prime }=\left|\begin{array}{ccc}b& c& a+b\\ a& b& c+a\\ c& a& b+c\end{array}\right|+\left|\begin{array}{ccc}b& a& a+b\\ a& c& c+a\\ c& b& b+c\end{array}\right|+\left|\begin{array}{ccc}c& c& a+b\\ b& b& c+a\\ a& a& b+c\end{array}\right|+\left|\begin{array}{ccc}c& a& a+b\\ b& c& c+a\\ a& b& b+c\end{array}\right|$
$D^{\prime }=\left|\begin{array}{ccc}b& c& a+b\\ a& b& c+a\\ c& a& b+c\end{array}\right|+\left|\begin{array}{ccc}b& a& a+b\\ a& c& c+a\\ c& b& b+c\end{array}\right|+\left|\begin{array}{ccc}c& a& a+b\\ b& c& c+a\\ a& b& b+c\end{array}\right|$ (Since,a determinant with two identical rows/columns has a value 0)
$D^{\prime }=\left|\begin{array}{ccc}b& c& a\\ a& b& c\\ c& a& b\end{array}\right|+\left|\begin{array}{ccc}b& c& b\\ a& b& a\\ c& a& c\end{array}\right|+\left|\begin{array}{ccc}b& a& a\\ a& c& c\\ c& b& b\end{array}\right|+\left|\begin{array}{ccc}b& a& b\\ a& c& a\\ c& b& c\end{array}\right|+\left|\begin{array}{ccc}c& a& a\\ b& c& c\\ a& b& b\end{array}\right|+\left|\begin{array}{ccc}c& a& b\\ b& c& a\\ a& b& c\end{array}\right|$
$D^{\prime }=\left|\begin{array}{ccc}b& c& a\\ a& b& c\\ c& a& b\end{array}\right|+\left|\begin{array}{ccc}c& a& b\\ b& c& a\\ a& b& c\end{array}\right|$
$=\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|+\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$ (On interchanging rows , determinant value changes by -sign)
$\Rightarrow D^{\prime }=2D$