Question

The value of x, so that the matrix $=\begin{bmatrix} x+a &b& c \\[0.3em] a& x+b & c \\[0.3em] a & b & x+c \end{bmatrix}$ has rank 3 , is

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

Answer 1

Answer: (C)

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