Answer
$z=(\frac{3-17i}{2},\frac{-3+3i}{2},\frac{-9+i}{2},\frac{27+23i}{2})$
Work Step by Step
Given: $2x+3iy=(1+i)z$
Plug in the given values:
$2(5+i,0,−1−2i,1+8i)+3i(−3,i,i,3)=(1+i)z$
$(10+2i,0,-2-4i,2+16i)+(-9i,-3,-3,9i)=(1+i)z$
$(10-7i,-3,-5-4i,2+25i)=(1+i)z$
Hence, $z=\frac{1-i}{2}(10-7i,-3,-5-4i,2+25i)$
$z=(\frac{10-7i-10i-7}{2},\frac{-3+3i}{2},\frac{-5-4i+5i-4}{2},\frac{2+25i-2i+25}{2}$
$z=(\frac{3-17i}{2},\frac{-3+3i}{2},\frac{-9+i}{2},\frac{27+23i}{2})$
