Question

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

• A. $ y\ne (a-b+c) $
• B. $ y\ne 1 $
• C. $ y=0 $
• D. $ y\ne -(a+b+c) $ and $ y\ne 0 $

Answer 1

Answer: (D)

Here, the rank of $A$ is $3$ .
Therefore, the minor of order $3$ of $A \neq 0$
$\Rightarrow \begin{vmatrix}y+a & b & c \\ a & y+b & c \\ a & b & y+c\end{vmatrix} \neq 0$
[Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, and taking $(y+a=b+c)$ common from $C_{1}$ ]
$\Rightarrow (y+a+b+c)\begin{vmatrix}1 & b & c \\ 1 & y+b & c \\ 1 & b & y+c\end{vmatrix} \neq 0$
[Applying $\left.R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}\right]$
$\Rightarrow (y+a+b+c)\begin{vmatrix}1 & b & c \\ 0 & y & 0 \\ 0 & 0 & y\end{vmatrix}\neq 0$
Expanding along $C_{1}$
$\Rightarrow (y+a+b+c)\left(y^{2}\right) \neq 0$
$\Rightarrow y \neq 0$ and $y \neq-(a+b+c)$