Question

If the matrix $A\left|\begin{array}{ccc}y+a& b& c\\ a& y+b& c\\ a& b& y+c\end{array}\right|$ 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 $A\ne 0$
$\Rightarrow \left|\begin{array}{ccc}y+a& b& c\\ a& y+b& c\\ a& b& y+c\end{array}\right|\ne 0$
[Applying $C_1\to C_1+C_2+C_3$, and taking $(y+a=b+c)$ common from $C_1$ ]
$\Rightarrow (y+a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& y+b& c\\ 1& b& y+c\end{array}\right|\ne 0$
[Applying $R_2\to R_2-R_1,R_3\to R_3-R_1]$
$\Rightarrow (y+a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& y& 0\\ 0& 0& y\end{array}\right|\ne 0$
Expanding along $C_1$
$\Rightarrow (y+a+b+c)\left(y^2\right)\ne 0$
$\Rightarrow y\ne 0$ and $y\ne -(a+b+c)$