A) \[{{b}_{0}}=0,{{b}_{1}}={{n}^{2}}-3n+3\]
B) \[\left| \begin{matrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \\ \end{matrix} \right|\]
C) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-3abc\]
D) \[3ab\]
Correct Answer: D
Solution :
\[\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1+y \\ \end{matrix} \right|=\left| \begin{matrix} 1 & 0 & 0 \\ 1 & -x & 1 \\ 1 & 0 & y \\ \end{matrix} \right|\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]\[=-xy\]
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