(C) Given the determinant: $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$ Expanding the determinant: $(1-i)(-i) - (i)(1+2i) =

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(C) Given the determinant: $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$ Expanding the determinant: $(1-i)(-i) - (i)(1+2i) = x+iy$ $-i + i^2 - (i + 2i^2) = x+iy$ Since $i^2 = -1$,we substitute: $-i - 1 - (i - 2) = x+iy$ $-i - 1 - i + 2 = x+iy$ $1 - 2i = x+iy$ Comparing the real and imaginary parts,we get $x = 1$ and $y = -2$.

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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If $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$,then $x$ is equal to

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$-2$
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$-1$
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$1$
D
$2$

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If $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$,then $x$ is equal to

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