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

Answer 1

Answer: (D)

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

$\left[\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right]=0$

Apply $C_1\to C_1+C_2+C_3$

$\left[\begin{array}{ccc}x+a+b+c& b& c\\ x+a+b+c& x+b& c\\ x+a+b+c& b& x+c\end{array}\right]=0$

Take $x+a+b+c$ as a common factor from $C_1$

$(x+a+b+c)\left[\begin{array}{ccc}1& b& c\\ 1& x+b& c\\ 1& b& x+c\end{array}\right]=0$

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

$(x+a+b+c)\left[\begin{array}{ccc}1& b& c\\ 0& x& 0\\ 0& 0& x\end{array}\right]=0$

Expanding the above matrix along $C_1$, we have

$(x+a+b+c)[1(x^2-0)-0+0]=0$

$\Rightarrow (x+a+b+c)x^2=0$

$\Rightarrow (x+a+b+c)=0\text{or}{x}^2=0$

$\Rightarrow x=-(a+b+c)\text{or}x=0$

Hence, for $\left[\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right]=0$,

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