Question

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

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

Answer 1

The correct option is D $x=0,y=0$

$\left|\begin{array}{ccc}6i& -3i& 1\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|$

Determinant by first row

$6i\left|\begin{array}{cc}3i& -1\\ 3& i\end{array}\right|-\left(3i\right)\left|\begin{array}{cc}4& -1\\ 20& i\end{array}\right|+\left(1\right)\left|\begin{array}{cc}4& 3i\\ 20& 3\end{array}\right|$

$=6i(3i\times i-(-1)\times 3)+3i(4i-(-1)\times 20)+1(12-3i\times 20)$

$=6i(-3+3)+3i(4i+20)+12-60i$

$=6i\times 0+12(i{)}^2+60i+12-60i$

$=0-12+60i+12-60i$

$=0+0i$

on comparing we get

$x=0$ & $y=0$