A) (3, 1)
B) (1, 3)
C) (0, 3)
D) (0, 0)
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} 6i\,\,\, & -3i\,\,\, & \,\,1 \\ 4\,\, & 3i & -1 \\ 20\, & 3 & \,\,i \\ \end{matrix}\, \right|\]=\[x+iy\] Þ \[\left| \,\begin{matrix} 6i+4\,\,\,\, & 0\,\,\,\, & \,\,0 \\ 4\,\,\, & 3i\,\,\,\, & -1 \\ 20\,\,\, & 3\,\,\,\, & \,\,i \\ \end{matrix}\, \right|=x+iy\] \[[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}]\] Þ \[(6i+4)\,(3{{i}^{2}}+3)\]= \[x+iy\] Þ \[(6i+4)\,(-3+3)=x+iy\] Þ \[x+iy=0\,\,=0+i.0\] Þ \[(x,\,y)\,=(0,\,0)\].You need to login to perform this action.
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