A) \[16\]
B) \[25\]
C) \[27\]
D) \[36\]
Correct Answer: B
Solution :
[b] \[|A|\,=\left| \begin{matrix} a & 2 & 1 \\ 0 & b & 0 \\ 0 & -3 & c \\ \end{matrix} \right|=abc=1\] Also, \[{{x}^{3}}-3{{x}^{2}}+2x-1=(x-a)(x-b)(x-c)\] ?(1) Now, \[{{A}^{2}}-{{B}^{2}}=4I-4B\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{A}^{2}}-({{B}^{2}}-4B+4I)=O\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{A}^{2}}-{{(B-2I)}^{2}}=O\] \[\Rightarrow \,\,\,\,\,\,\,\,\,(A+B-2I)\,\,(A-B+2I)=O\] \[\because \,\,\,\,\,\,\,\,\,A+B-2I\ne O,\,\,\,\,\therefore A-B+2I=O\] \[\therefore \,\,\,\,\,\,\,\,\,\,B=A+2I=\left| \begin{matrix} a+2 & 2 & 1 \\ 0 & b+2 & 0 \\ 0 & -3 & c+2 \\ \end{matrix} \right|\] \[\therefore \,\,\,\,\,\,\,\,\left| B \right|=(a+2)\,\,(b+2)\,(c+2)\] ?.(2) Putting \[x=-2\] into (1), we get \[-8-12-4-1=(-2-a)\,(-2-b)(-2-c)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,25=(a+2)\,(b+2)\,(c+2)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\left| B \right|=25\]
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