Question

Prove that
$\left|\begin{array}{ccc}b+c-a-d& bc-ad& bc(a+d)-ad(b+c)\\ c+a-b-d& ca-bd& ca(b+d)-bd(c+a)\\ a+b-c-d& ab-cd& ab(c+d)-cd(a+b)\end{array}\right|=-2(b-c)(c-a)(a-b)(a-d)(b-d)(c-d)$.

Answer 1

Given $\left|\begin{array}{ccc}b+c-a-d& bc-ad& bc(a+d)-ad(b+c)\\ c+a-b-d& ca-bd& ca(b+d)-bd(c+a)\\ a+b-c-d& ab-cd& ab(c+d)-cd(a+b)\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}2(c-a)& (b+d)(c-a)& 2bd(c-a)\\ 2(c-b)& (a+d)(c-b)& 2ad(c-b)\\ 2(a-d)& (b+c)(a-d)& 2bc(a-d)\end{array}\right|[R_1\to R_1–R_3;R_2\to R_2–R_3;2R_3\to 2R_3+R_2]$

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

$=2(c-a)(c-b)(a-d)\left|\begin{array}{ccc}1& b+d& bd\\ 0& a-b& d(a-b)\\ 0& c-d& b(c-d)\end{array}\right|[R_2\to R_2-R_1;R_3\to R_3-R_1]$

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

$=2(c-a)(c-b)(a-d)(a-b)(c-d)(b-d)$

$=-2(b-c)(c-a)(a-b)(a-d)(b-d)(c-d)$ $\left[henceproved\right]$