Question

Using properties of determinants, show that $\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|=4abc.$

Answer 1

Let $det\left(A\right)=\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$=(b+c)\left[\right(c+a\left)\right(a+b)-bc]-a\left[b\right(a+b)-bc]+a[bc-c(c+a\left)\right]$

$=(b+c)(ac+a^2+ab)-a^{2}b-a{b}^2+abc+abc-a{c}^2-a^{2}c$

$=abc+a^{2}b+a{b}^2+a{c}^2+a^{2}c+abc-a^{2}b-a{b}^2+abc+abc-a{c}^2-a^{2}c$

$=4abc$