Question

lf $\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|=\lambda \left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$ then $\lambda$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-3$

Answer 1

The correct option is B $-1$

$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|=\lambda \left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$

Let $A=\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|andB=\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$

By row transformation on A,

$R_1\to R_1-R_2$ and $R_3\to R_3-R_2$

Let $A=\left|\begin{array}{ccc}0& a-b& (b-a)c\\ 1& b& ca\\ 0& c-b& a(b-c)\end{array}\right|=(b-c)(b-a)\left|\begin{array}{ccc}0& -1& c\\ 1& b& ac\\ 0& -1& a\end{array}\right|=(b-c)(b-a)[-1(-a+c\left)\right]$

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

and

$B=\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$

By column transformation of B,

$C_1\to C_1-C_2$ and $C_3\to C_3-C_2$

$B=\left|\begin{array}{ccc}{a}^2-b^2& b^2& c^2-b^2\\ a-b& b& c-b\\ 0& 1& 0\end{array}\right|=(a-b)(c-b)\left|\begin{array}{ccc}a+b& b^2& c+b\\ 1& b& 1\\ 0& 1& 0\end{array}\right|$

$=(a-b)(c-b)(-1(a+b-(c+b)\left)\right)$

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

given $|A|=\lambda |B|$

$\Rightarrow (a-b)(b-c)(c-a)=\lambda (a-b)(b-c)(a-c)$

$\Rightarrow \lambda =-1$