Question

Show 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

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

$R_2\to R_2-2R_1,R_3\to R_3-2R_1$

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

Expanding along first column

$=a\left[a\right(7a+3b)-3a(2a+b\left)\right]-0+0$

$=a[7a^2+3ab-6a^2-3ab]$

$=a(a{)}^2=a^3=R.H.S.$