Answer:
Option C
Explanation:
A= $\begin{bmatrix}x & x&x \\x & x^{2}&x\\x&x&x+1 \end{bmatrix}$
Since , rank A=1 , therefore atleast one determined of order 1 should be non-zero and all the determinants of order 2 and 3 should be zero.
If x=0 , then $A=\begin{bmatrix}0 & 0&0 \\0 & 0&0\\0&0&1 \end{bmatrix}$ , which have non-zero
determinant of order 1 only.
If x=1, then $A=\begin{bmatrix}1 & 1&1 \\1 & 1&1\\1&1&2 \end{bmatrix}$, which have a non-zero determinant of order 2,
nearly $\begin{bmatrix}1 & 1 \\1 & 2 \end{bmatrix}$
$\therefore$ x can take value 0 only
