Question

The value of $\left|\begin{array}{c}1\\ bc\\ b+c\end{array}\begin{array}{c}1\\ ca\\ c+a\end{array}\begin{array}{c}1\\ ab\\ a+b\end{array}\right|$ is:

• A. $1$
• B. $0$
• C. $(a-b)(b-c)(c-a)$
• D. $(a+b)(b+c)(c+a)$

Answer 1

The correct option is C $(a-b)(b-c)(c-a)$

We need to find value of $\left|\begin{array}{ccc}1& 1& 1\\ bc& ca& ab\\ b+c& c+a& a+b\end{array}\right|$

Vertical lines here shows that this is a finding determinant

$=1\left|\begin{array}{cc}ca& ab\\ c+a& a+b\end{array}\right|-1\left|\begin{array}{cc}bc& ab\\ b+c& a+b\end{array}\right|+1\left|\begin{array}{cc}bc& ca\\ b+c& c+a\end{array}\right|$

$=1\left(ca\right(a+b)-ab(c+a\left)\right)-\left(1\right(bc(a+b)-ab(b+c))+1(bc(c+a)-ac(b+c))$

$=c{a}^2-a^{2}b-b^{2}c+a{b}^2+b{c}^2-a{c}^2$

$=(a-b)(b-c)(c-a)$