If `|6i-3i1 4 3i-1 20 3i|=x+i y ,` then a.`x=3,y=1` b. `x=1,y=3` c. `x=0,y=3` d. `x=0,y=0`

If |[6i,-3i,1],[ 4, 3i,-1],[ 20, 3,i]|=x+i y , then a. x=3,y=1 b. x=1,y=3 c. x=0,y=3 d. x=0,y=0

If |[6i,-3i,1],[ 4, 3i,-1],[ 20, 3,i]|=x+i y , then a. x=3,y=1 b. x=1,y=3 c. x=0,y=3 d. x=0,y=0

If det|6i -3i 1 4 3i-1 20 3 i|=x+i y , then a. x=3,y=1 b. x=1,y=3 c. x=0,y=3 d. x=0,y=0

If |[6i,-3i,1],[ 4, 3i,-1],[ 20, 3,i]|=x+i y , then

If |6i-3i143i-1203i|=x+iy, then a.x=3,y=1 b.x=1,y=3 c.x=0,y=3 d.x=0,y=0

\begin{bmatrix}6i&-3i&1\\ 4 &3i& -1\\ 20&3&i\\ \end{bmatrix} =x+i y , then a. x=3,y=1 b. x=1,y=3 c. x=0,y=3 d. x=0,y=0

If |[6i,-3i,1],[4,3i,-1],[20,3,i]| = x+iy, x, y in R then (x+y) =

If |{:(6i , -3i , 1) , (4 , 3i , -1) , (20 , 3 , i):}| = x +iy show that x = y = 0

The real values of xa n dy for which the following equation is satisfied: ((1+i)(x-2i))/(3+i)+((2-3i)(y+i))/(3-i)=i x=3,y=1 b. x=3, y=-1 c. x=-3,y=1 d. x=-3,y=-1