A) \[y\ne (a-b+c)\]
B) \[y\ne 1\]
C) \[y=0\]
D) \[y\ne -(a+b+c)\]and\[y\ne 0\]
Correct Answer: D
Solution :
Here, the rank of\[A\]is 3. Therefore, the minor of order 3 of\[A\ne 0\] \[\Rightarrow \] \[\left| \begin{matrix} y+a & b & c \\ a & y+b & c \\ a & b & y+c \\ \end{matrix} \right|\ne 0\] [Applying\[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\], and taking\[(y+a=b+c)\]common from\[{{C}_{1}}\]] \[\Rightarrow \] \[(y+a+b+c)\left| \begin{matrix} 1 & b & c \\ 1 & y+b & c \\ 1 & b & y+c \\ \end{matrix} \right|\ne 0\] [Applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}]\] \[\Rightarrow \] \[(y+a+b+c)\left| \begin{matrix} 1 & b & c \\ 0 & y & 0 \\ 0 & 0 & y \\ \end{matrix} \right|\ne 0\] Expanding along \[{{C}_{1}}\] \[\Rightarrow \,\,\,(y+a+b+c)\,({{y}^{2}})\ne 0\] \[\Rightarrow \,\,y\ne 0\] and \[y\ne -(a+b+c)\]
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