Question

If a, b, c and d are complex numbers, then the determinant
$\Delta =\left|\begin{array}{ccc}2& a+b+c+d& ab+cd\\ a+b+c+d& 2(a+b)(c+d)& ab(c+d)+cd(a+b)\\ ab+cd& ab(c+d)+cd(a+b)& 2abcd\end{array}\right|$
equals

• A. 0
• B. abcd
• C. a+b+c+d
• D. 2abcd

Answer 1

The correct option is A 0

$\Delta =\left|\begin{array}{ccc}2& a+b+c+d& ab+cd\\ a+b+c+d& 2(a+b)(c+d)& ab(c+d)+cd(a+b)\\ ab+cd& ab(c+d)+cd(a+b)& 2abcd\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 0\\ c+d& a+b& 0\\ cd& ab& 0\end{array}\right|\left|\begin{array}{ccc}1& 1& 0\\ a+b& c+d& 0\\ ab& cd& 0\end{array}\right|$ Row X Row multiplication
$=0$
Hence, option A.