Question

$\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3a\end{array}\right|=a^3$

Answer 1

$\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3a\end{array}\right|$

$R_2\to R_2-2R_1$

$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a-2a& 3a+2b-2a-2b& 4a+3b+2c-2a-2b-2c\\ 3a& 6a+3b& 10a+6b+3a\end{array}\right|$

$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2c+b\\ 3a& 6a+3b& 10a+6b+3a\end{array}\right|$

$R_3\to R_3-3R_1$

$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2c+b\\ 3a-3a& 6a+3b-3a-3b& 10a+6b+3c-3a+b+c\end{array}\right|$

$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2c+b\\ 0& 6a+3b-3a-3b& 10a+6b+3c-3a+b+c\end{array}\right|$

$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2c+b\\ 0& 3a& 7a+3b\end{array}\right|$

Expanding along $C_1$

$=a\left|\begin{array}{cc}a& 2a+b\\ 3a& 7a+3b\end{array}\right|-0+0$

$=a(7a^2+3ab-6a^2-3ab)$

$=a^3$

$=$R.H.S

Hence proved.