Matrix and Determinant
79108
If \(\left|\begin{array}{ccc}3 i & -9 i & 1 \\ 2 & 9 i & -1 \\ 10 & 9 & i\end{array}\right|=x+i y\), then
1 \(x=1, y=1\)
2 \(x=0, y=1\)
3 \(\mathrm{x}=1, \mathrm{y}=0\)
4 \(x=0, y=0\)
5 \(\mathrm{x}=-1, \mathrm{y}=0\)
Explanation:
(D) : It is given that,
\(\left|\begin{array}{ccc} 3 \mathrm{i} & -9 \mathrm{i} & 1 \\ 2 & 9 \mathrm{i} & -1 \\ 10 & 9 & \mathrm{i} \end{array}\right|=\mathrm{x}+\mathrm{iy}\)
Applying \(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}\)
\(\left|\begin{array}{ccc} 3 i+2 & 0 & 0 \\ 2 & 9 i & -1 \\ 10 & 9 & i \end{array}\right|=x+i y\)
\((3 i+2)\left[9 i^{2}+9\right]=x+i y\)
\((3 i+2)(-9+9)=x+i y\)
\(0=x+i y\)
\(x=0, y=0\)
\(\left(\because \mathrm{i}^{2}=-1\right)\)