Question

if $a,b,c$ are unequal what is the condition that the value of following determinat is zero $\delta =\left|\begin{array}{ccc}a& a^2& a^3+1\\ b& b^2& b^3+1\\ c& c^2& c^3+1\end{array}\right|$

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

Answer 1

Answer: (C)

$(a-b)(b-c)(c-a)=0$

Solution:- (C) $(a-b)(b-c)(c-a)=0$

$\delta =\left|\begin{array}{ccc}a& a^2& a^3+1\\ b& b^2& b^3+1\\ c& c^2& c^3+1\end{array}\right|$

Apply $C_3\to C_3-(C_2\times C_1)$

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

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

$\Rightarrow \delta =\left|\begin{array}{ccc}a& a^2& 1\\ b-a& b^2-a^2& 0\\ c-a& c^2-a^2& 0\end{array}\right|$

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

Taking $(b-a)$ and $(c-a)$ as common factor from $C_2$ and $C_3$ respectively.

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

Expanding the above determinant along $C_3$, we have

$\delta =(b-a)(c-a)[1((c+a)-(b+a))-0+0]$

$\Rightarrow \delta =(b-a)(c-a)(c+a-b-a)$

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

Hence, for $\delta =0$,

$(b-a)(c-a)(c-b)=0$

$\Rightarrow (a-b)(b-c)(c-a)=0$.