Question

Using properties of determinants, 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

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

Operating $R_2\to R_2-2R_1$ and $R_3-3R_1$, we get

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

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

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

$=a.a^2=a^3$