If `A=[{:(4i-6,10i),(14i,6+4i):}]and k(1)/(ki), where i- sqrt(-1),` then kA is equal to

Options

A. `[{:(2+3i,5),(7,2-3i):}]`
B. `[{:(2-3i,5),(7,2+3i):}]`
C. `[{:(2-3i,7),(5,2+3i):}]`
D. `[{:(2+3i,5),(7,2+3i):}]`

Answer

A

Explanation

`A= [{:(4i6,10i),(14i, 6+4i):}]and K=(1)/(2i)`
` K=(1)/(2i)=(i)/(2i(i))=(-i)/(2).`
`thereforeKA=(-i)/(2)[{:(4i-6,10i),(14i,6+4i):}]`
`=[{:((4i-6)((-i)/(2)),10i((-i)/(2))),(14i((-i)/(1-i)),(6+4i)((-i)/(2))):}]`
`=[{:(-2i^(2)+3i,-5i^(2)),(-7i^(2),-3i-2i^(2)):}]=[{:(2+3i,5),(7,2-3i):}]`