Question

If $\left|\begin{array}{ccc}6\mathrm{i}& -3\mathrm{i}& 1\\ 4& 3\mathrm{i}& -1\\ 20& 3& \mathrm{i}\end{array}\right|=\mathrm{x}+\mathrm{iy}$, then

• A. $x=3,y=1$
• B. $x=3,y=0$
• C. $x=0,y=1$
• D. $x=0,y=0$

Answer 1

The correct option is D

$x=0,y=0$

Explanation for the correct option:

Finding the value of $x$ and $y$:

Given that,

$\left|\begin{array}{ccc}6\mathrm{i}& -3\mathrm{i}& 1\\ 4& 3\mathrm{i}& -1\\ 20& 3& \mathrm{i}\end{array}\right|=\mathrm{x}+\mathrm{iy}$

Expanding the determinant along $R_1$, we get

$\begin{array}{rcl}\left|\begin{array}{ccc}6\mathrm{i}& -3\mathrm{i}& 1\\ 4& 3\mathrm{i}& -1\\ 20& 3& \mathrm{i}\end{array}\right|& =& 6i\left(3i^2+3\right)-\left(-3i\right)\left(4i+20\right)+\left(1\right)\left(12-60i\right)\\ & =& 18i^3+18i+12i^2+60i+12-60i\\ & =& -i18+18i-12+12\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{]}\\ & =& 0+i\left(0\right)\end{array}$

Now, by comparing with $x+iy$ we get,

$x=0$ and $y=0$

Hence, the correct option is D.