Question

A = $\left[\begin{array}{ccc}-qr& p(q+r)& pr+pq\\ pq+qr& -pr& pq+qr\\ qr+pr& qr+pr& -pq\end{array}\right]$ then $\left|A\right|$ equals:

• A. ${\left(\sum pq\right)}^2$
• B. ${\left(\sum p^2{q}^2\right)}^2$
• C. $\left[\sum (qr{)}^3\right]$
• D. ${\left(\sum pq\right)}^3$

Answer 1

The correct option is D ${\left(\sum pq\right)}^3$

$A$ = $\left[\begin{array}{ccc}-qr& p(q+r)& pr+pq\\ pq+qr& -pr& pq+qr\\ qr+pr& qr+pr& -pq\end{array}\right]$
$|A|$ = $\left|\begin{array}{ccc}-qr& p(q+r)& pr+pq\\ pq+qr& -pr& pq+qr\\ qr+pr& qr+pr& -pq\end{array}\right|$
applying $R_1\to R_1+R_2+R_3$ gives
$\to (pq+qr+pr)\left|\begin{array}{ccc}1& 1& 1\\ pq+qr& -pr& pq+qr\\ qr+pr& qr+pr& -pq\end{array}\right|$
applying $C_1\to C_1-C_3$ and expanding gives
$\to (pq+qr+pr)\left|\begin{array}{ccc}0& 1& 1\\ 0& -pr& pq+qr\\ pq+qr+pr& qr+pr& -pq\end{array}\right|={(pq+qr+pr)}^3$
$\therefore |A|={(pq+qr+pr)}^3$
Hence, option D.