Question

Prove that:
$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|={(a+b+c)}^3$

Answer 1

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

We use column transformations:

1. Column 1 = Column 1 + Column 2 + Column 3

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

$=2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|$

$=2(a+b+c){\Delta }_1$

where:

${\Delta }_1=\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|$

We use elementary row transformations on this new determinant:

2. Row 2 = Row 2 - Row 1 and Row 3 = Row 3 - Row 1

${\Delta }_1=\left|\begin{array}{ccc}1& a& b\\ 0& b+c+a& 0\\ 0& 0& c+a+b\end{array}\right|$

3. Expanding this determinant along Column 1, we get:

${\Delta }_1=(a+b+c{)}^2$

Thus, we get:

$\Delta =2(a+b+c){\Delta }_1=2(a+b+c{)}^3$

Hence proved.