Question

If $\left|\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right|=0$, then $x$ equals

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

Answer 1

Answer: (B)

$0,-(a+b+c)$

Given $\left|\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right|=0$
$C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}x+a+b+c& b& c\\ x+a+b+c& x+b& c\\ x+a+b+c& b& x+c\end{array}\right|=0$
Taking common $(x+a+b+c)$ from $C_1$

$(x+a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& x+b& c\\ 1& b& x+c\end{array}\right|=0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$(x+a+b+c)\left|\begin{array}{ccc}0& -x& 0\\ 0& x& -x\\ 1& b& x+c\end{array}\right|=0$
$\Rightarrow x^2(x+a+b+c)=0$

$\Rightarrow x=0,-(a+b+c)$