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)

$x\neq\,0\,$ and $\,x\neq -(a+b+c)$

Answer 2

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 $x \neq 0, x + a + b + c \neq 0$ $i.e., x \neq -(a + b + c)$