Question

Prove that $\left|\begin{array}{cccc}a& b& c& d\\ b& a& d& c\\ c& d& a& b\\ d& c& b& a\end{array}\right|=(a+b+c+d)(a-b+c-d)(a-b-c+d)(a+b-c-d)$.

Answer 1

Given $\left|\begin{array}{cccc}a& b& c& d\\ b& a& d& c\\ c& d& a& b\\ d& c& b& a\end{array}\right|$

$=(a+b+c+d)\left|\begin{array}{cccc}1& 1& 1& 1\\ b& a& d& c\\ c& d& a& b\\ d& c& b& a\end{array}\right|[R_1\to R_1+R_2+R_3+R_4]$

$=(a+b+c+d)\left|\begin{array}{cccc}1& 0& 0& 0\\ b& a-b& d-b& c-b\\ c& d-c& a-c& b-c\\ d& c-d& b-d& a-d\end{array}\right|[C_2\to C_2-C_1;C_3\to C_3-C_1;C_4\to C_4-C_1]$

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

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

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

$=(a+b+c+d)(a+c-b-d)(a+b-c-d)\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ c-d& b-d& a-c\end{array}\right|[C_3\to C_3–C_1]$

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