Question

$\left|\begin{array}{ccc}-2a& a+b& a+c\\ b+a& -2a& b+c\\ c+a& c+b& -c\end{array}\right|=$

• A. $-4(b+c)(c+a)(a+b)$
• B. $4(b+c)(c+a)(a+b)$
• C. $\sqrt{4(b+c)(c+a)(a+b)}$
• D. $2(b+c)(c+a)(a+b)$

Answer 1

Answer: (B)

$4(b+c)(c+a)(a+b)$

Let $\Delta =\left|\begin{array}{ccc}-2a& a+b& a+c\\ b+a& -2a& b+c\\ c+a& c+b& -c\end{array}\right|=$
Putting $a+b=0$
$\Rightarrow b=-a$
then
$\Delta =\left|\begin{array}{ccc}-2a& 0& a+c\\ 0& 2a& c-a\\ c+a& c-a& -2c\end{array}\right|=$Expanding along $R_1$
$=-2a\{-4ac-(c-a{)}^2\}-0+(a+c\left)\right\{0-2a(c+a)\}$
$=2a(c+a{)}^2-2a(c+a{)}^2$
$=0$
Hence a + b is a factor of $\Delta$ similarly b + c and c + a are the factors of $\Delta$.
On
expansion of determinant we can see that each term of the determinant
is a homogeneous expression in a, b, c of degree 3 and also R.H.S. is a
homogeneous expression of degree 3.
Let $\Delta =k(a+b)(b+c)(c+a)$
or $\left|\begin{array}{ccc}-2a& a+b& a+c\\ b+a& -2a& b+c\\ c+a& c+b& -c\end{array}\right|=k(a+b)(b+c)(c+a)$
Putting $a=0,b=1,c=2$, we get
$\left|\begin{array}{ccc}0& 1& 2\\ 1& -2& 3\\ 2& 3& -4\end{array}\right|=k(0+1)(1+2)(2+0)$
$\Rightarrow 0-1(-4-6)+2(3+4)=6k$
$\Rightarrow 24=6k$
$\therefore k=4$
Hence $\left|\begin{array}{ccc}-2a& a+b& a+c\\ b+a& -2a& b+c\\ c+a& c+b& -c\end{array}\right|=4(a+b)(b+c)(c+a)$