# If the matrix $A=\begin{bmatrix}y+a&b&c\\a&y+b&c\\a&b&y+c\end{bmatrix}$ has rank 3 then

$\begin{array}{1 1}(a)\;y\neq (a+b+c)&(b)\;y\neq 1\\(c)\;y=0&(d)\;y\neq -(a+b+c)\;and\;y\neq 0\end{array}$

Here the rank of A is 3.
$\therefore$ minor of order 3 of $A\neq 0$
$\Rightarrow \begin{vmatrix}y+a&b&c\\a&y+b&c\\a&b&y+c\end{vmatrix}\neq 0$
Apply $C_1\rightarrow C_1+C_2+C_3$ and taking $(y+a+b+c)$ common from $C_1$
Apply $R_2\rightarrow R_2-R_1$
$R_3\rightarrow R_3-R_1$ we get
$\Rightarrow (y+a+b+c)\begin{vmatrix}1&b&c\\0&y&0\\0&0&y\end{vmatrix}\neq 0$
expanding along $C_1$ we get
$\Rightarrow (y+a+b+c)(y^2)\neq 0\rightarrow$
$\Rightarrow y\neq 0$ and $y\neq -(a+b+c)$
Hence (d) is the correct answer.
edited Mar 19, 2014