Question

If $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$,
where $a,b,c$ are all different,then the determinant
$\left|\begin{array}{ccc}1& 1& 1\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ (x-b)(x-c)& (x-c)(x-a)& (x-a)(x-b)\end{array}\right|$
vanishes when

• A. $x=a+b-c$
• B. $x=\frac{1}{3}(a+b+c)$
• C. $x=\frac{1}{2}(a+b+c)$
• D. $x=a+b+c$

Answer 1

The correct option is B $x=\frac{1}{3}(a+b+c)$

The given determinant is $\left|\begin{array}{ccc}1& 1& 1\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ (x-b)(x-c)& (x-c)(x-a)& (x-a)(x-b)\end{array}\right|$

Multiplying column 1, column 2 and column 3 by $(x-a)$, $(x-b)$ and $(x-c)$ respectively, we get

$\frac{1}{(x-a)(x-b)(x-c)}\left|\begin{array}{ccc}x-a& x-b& x-c\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ (x-a)(x-b)(x-c)& (x-b)(x-c)(x-a)& (x-a)(x-b)(x-c)\end{array}\right|$

Taking $(x-a)(x-b)(x-c)$ common from third row, we get

$\left|\begin{array}{ccc}x-a& x-b& x-c\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ 1& 1& 1\end{array}\right|$

Using row transformation, the above determinant can be written as

$\left|\begin{array}{ccc}1& 1& 1\\ x-a& x-b& x-c\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\end{array}\right|$

As per the question, the above determinant can be written as

$(x-a-(x-b\left)\right)(x-b-(x-c\left)\right)(x-c-(x-a\left)\right)\left(\right(x-a)+(x-b)+(x-c\left)\right)$

$=(b-a)(c-b)(a-c)(3x-a-b-c)$

The above expression is zero if $3x-a-b-c=0$ ($\because a\ne b\ne c$)

$\Rightarrow x=\frac{1}{3}(a+b+c)$