A) \[x=3,\,y=1\]
B) \[x=1,\,y=3\]
C) \[x=0,\,y=3\]
D) \[x=0,\,y=0\]
Correct Answer: D
Solution :
Now, \[\left| \begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix} \right|\] Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\] \[=\left| \begin{matrix} 6i+4 & 0 & 0 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix} \right|\] \[=(6i+4)\,(3{{i}^{2}}+3)\] = 0 But \[\left| \begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix} \right|=x+i\,y\] \[\Rightarrow \] \[0+0i=x+i\,y\] \[\Rightarrow \] \[x=0,\,\,y=0\]
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