A) 2
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
\[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\,+{{C}_{3}};\] we get \[=(a+b+c+s)\,\left| \begin{matrix} 1 & a & b \\ 1 & s+a & b \\ 1 & a & s+b \\ \end{matrix} \right|\] \[=2s\,\left| \begin{matrix} 0 & -s & 0 \\ 0 & s & -s \\ 1 & a & s+b \\ \end{matrix} \right|\,=2{{s}^{3}}=54\] \[\therefore \,\,{{s}^{3}}=27\] \[s=3\]You need to login to perform this action.
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