Question

If $\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|=16$, then $\left|\begin{array}{ccc}p+x& a+x& a+p\\ q+y& b+y& b+q\\ r+z& c+z& c+r\end{array}\right|$ is equal to

• A. $30$
• B. $32$
• C. $34$
• D. $36$

Answer 1

The correct option is B $32$

Let $A=\left|\begin{array}{ccc}p+x& a+x& a+p\\ q+y& b+y& b+q\\ r+z& c+z& c+r\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$, we get,

$A=\left|\begin{array}{ccc}2(p+a+x)& a+x& a+p\\ 2(q+b+y)& b+y& b+q\\ 2(r+c+z)& c+z& c+r\end{array}\right|$

$=2\left|\begin{array}{ccc}p+a+x& a+x& a+p\\ q+b+y& b+y& b+q\\ r+c+z& c+z& c+r\end{array}\right|$

Applying $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$, we get,

$A=2\left|\begin{array}{ccc}p& x-p& a+p\\ q& y-q& b+q\\ r& z-r& c+r\end{array}\right|$

Applying $C_2\to C_2+C_1$ and $C_3\to C_3-C_1$, we get,

$A=2\left|\begin{array}{ccc}p& x& a\\ q& y& b\\ r& z& c\end{array}\right|$

Applying $C_2\leftrightarrow C_3$ and then $C_1\leftrightarrow C_2$, we get,

$A=2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$

$A=2\times 16=32$