Question

The value of x, so that the matrix $=\left[\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right]$ has rank 3 , is

• A. $x\ne 0$
• B. x = a + b + c
• C. $x\ne 0$ and $x\ne -(a+b+c)$
• D. $x=0$ and $a=a+b+c.$

Answer 1

Answer: (C)

Since rank = 3.
$\therefore$ $\left|\begin{array}{ccc}x+a& b& c\\ a& a+b& c\\ a& b& x+c\end{array}\right|\ne 0$
Operate $C_1+C_2+C_3$
$\left|\begin{array}{ccc}x+a+b+c& b& c\\ x+a+b+c& x+b& c\\ x+a+b+c& b& x+c\end{array}\right|\ne 0$
$\Rightarrow$ $(x+a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& x+b& c\\ 1& b& x+c\end{array}\right|\ne 0$
$\Rightarrow$ $x+a+b+c\ne 0$ and
$\left|\begin{array}{ccc}1& b& c\\ 0& x& 0\\ 0& 0& x\end{array}\right|\ne 0\Rightarrow x^2\ne 0\Rightarrow x\ne 0$
Then $x\ne 0,x+a+b+c\ne 0$
$i.e.,x\ne -(a+b+c)$