Question

Prove that $\left|\begin{array}{ccc}{a}^2& bc& ac+c^2\\ a^2+ab& b^2& ac\\ ab& b^2+bc& c^2\end{array}\right|$ $=4a^2{b}^2{c}^2$

Answer 1

$\left|\begin{array}{ccc}{a}^2& bc& ac+c^2\\ a^2+ab& b^2& ac\\ ab& b^2+bc& c^2\end{array}\right|$

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

Taking $a,b,c$ common from $C_1,C_2,C_3$

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

$C_1\to C_1+C_2+C_3$

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

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

$C_1\to C_1-C_3$

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

Taking $b$ common from $C_1$

$=2a{b}^{2}c\left|\begin{array}{ccc}0& c& a+c\\ 1& b& a\\ 1& b+c& c\end{array}\right|$

$R_3\to R_3-R_2$

$=2a{b}^{2}c\left|\begin{array}{ccc}0& c& a+c\\ 1& b& a\\ 0& c& c-a\end{array}\right|$

Expanding along the first column, we get

$=2a{b}^{2}c[-1(c^2-ac-ac-c^2\left)\right]$

$=4a^2{b}^2{c}^2$