A) \[4abc\]
B) \[abc\]
C) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
D) \[4{{a}^{2}}bc\]
E) \[4{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
Correct Answer: A
Solution :
\[\left| \begin{matrix} a+b & a & b \\ a & a+c & c \\ b & c & b+c \\ \end{matrix} \right|=\left| \begin{matrix} b & -c & b-c \\ a & a+c & c \\ b & c & b+c \\ \end{matrix} \right|\] \[(By\,{{R}_{1}}\to {{R}_{1}}-{{R}_{2}})\] \[=\left| \begin{matrix} 2b & 0 & 2b \\ a & a+c & c \\ b & c & b+c \\ \end{matrix} \right|\] \[(by\,{{R}_{1}}\to {{R}_{1}}+{{R}_{3}})\] \[=\left| \begin{matrix} 2b & 0 & 0 \\ a & a+c & c-a \\ b & c & c \\ \end{matrix} \right|\] \[(By\,{{C}_{3}}\to {{C}_{3}}-{{C}_{1}})\] \[=2b(ac+{{c}^{2}}-{{c}^{2}}+ac)=4abc\]
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