Question

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

Answer 1

To prove:
$\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3c\end{array}\right|=a^3$
Taking L.H.S.
$\left|\begin{array}{ccc}a& a+b& a+b+c\\ 2a& 3a+2b& 4a+3b+2c\\ 3a& 6a+3b& 10a+6b+3c\end{array}\right|$
Applying $R_2\to R_2-2R_1$
$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2a+b\\ 3a& 6a+3b& 10a+6b+3c\end{array}\right|$
Applying $R_3\to R_3-3R_1$
$=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 0& a& 2a+b\\ 0& 3a& 7a+3b\end{array}\right|$
$=a\left|\begin{array}{cc}a& 2a+b\\ 3a& 7a+3b\end{array}\right|-0+0$
$=a\left(a\right(7a+3b)-3a(2a+b\left)\right)$
$=a(7a^2+3ab-6a^2-3ab)$
$=a\left(a^2\right)$
$=a^3$
$=R.H.S.$
Thus $L.H.S.=R.H.S.$
Hence proved.