Question

The value of $\left|\begin{array}{ccc}2& a+b+c+d& ab+cd\\ a+b+c+d& 2(a+b)(c+d)& ab(c+d)+cd(a+b)\\ ab+cd& ab(c+d)+cd(a+b)& 2abcd\end{array}\right|$

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

Given $\left|\begin{array}{ccc}2& a+b+c+d& ab+cd\\ a+b+c+d& 2(a+b)(c+d)& ab(c+d)+cd(a+b)\\ ab+cd& ab(c+d)+cd(a+b)& 2abcd\end{array}\right|$

To solve this type, consider $a=b=c=d=1$.

Now we have,

$=\left|\begin{array}{ccc}2& 4& 2\\ 4& 8& 4\\ 2& 4& 2\end{array}\right|$

$=2(8\times 2-4\times 4)-4(4\times 2-2\times 4)+2(4\times 4-8\times 2)$

$=0$