Question

The value of $\left|\begin{array}{ccc}(b+c{)}^2& ba& ac\\ ba& (c+a{)}^2& cb\\ ca& cb& (a+b{)}^2\end{array}\right|$ is:

• A. $2abc(a+b+c{)}^3$
• B. $abc(a+b+c{)}^2$
• C. $\left|\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|$
• D. $\left|\begin{array}{ccc}(b-c{)}^2& a^2& a^2\\ b^2& (c-a{)}^2& b^2\\ c^2& c^2& (a-b{)}^2\end{array}\right|$

Answer 1

Answer: (A), (C)

$\left|\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|$

Multiplying $R_1,R_2,R_3$ by a,b,c respectively and dividing by abc, we get

$\Delta =\frac{1}{abc}\times \left|\begin{array}{ccc}a(b+c{)}^2& b{a}^2& c{a}^2\\ a{b}^2& b(c+a{)}^2& c{b}^2\\ a{c}^2& b{c}^2& c(a+b{)}^2\end{array}\right|$

Taking a,b,c common from $C_1,C_2$ and $C_3$ respectively, we get
$\Delta =\left|\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|$
Now applying $C_2\to C_2-C_1,C_3\to C_3-C_1$ we get:

$\Delta =\left|\begin{array}{ccc}(b+c{)}^2& (a+b+c)(a-b-c)& (a-b-c)(a+b+c)\\ b^2& (c+a+b)(c+a-b)& 0\\ c^2& 0& (a+b+c)(a+b-c)\end{array}\right|$

$\Rightarrow \Delta =(a+b+c{)}^2\times \left|\begin{array}{ccc}(b+c{)}^2& a-b-c& a-b-c\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

Applying $R_1\to R_1-(R_2+R_3)$ and the taking 2 common from $R_1$, we get

$\Delta =2(a+b+c{)}^2\times \left|\begin{array}{ccc}bc& -c& -b\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$
Now, applying $C_2\to b{C}_2+C_1,C_3\to c{C}_3+C_1$ gives

$\Delta =\frac{2(a+b+c{)}^2}{bc}\times \left|\begin{array}{ccc}bc& 0& 0\\ b^2& b(c+a)& b^2\\ c^2& c^2& c(a+b)\end{array}\right|$
$\Rightarrow \Delta =\frac{2(a+b+c{)}^2}{bc}bc\left[\right(bc+ba\left)\right(ca+cb)-b^2{c}^2]$
$\Rightarrow \Delta =2(a+b+c{)}^2[bc(ac+bc+ab+a^2-bc)]$
$\Rightarrow \Delta =2abc(a+b+c{)}^3$