Answer 1

`|(4"i", "i"^3, 2"i"),(1, 3"i"^2, 4),(5, -3, "i")|` = x + iy

∴ `4"i"|(3"i"^2, 4),(-3, "i")|-"i"^3|(1, 4),(5, "i")| + 2"i"|(1, 3"i"^2),(5, -3)|` = x + iy

∴ 4i (3i³ + 12) – i³(i – 20) + 2i ( – 3 – 15i²) = x + iy

∴ 12i⁴ + 48i – i⁴ + 20i³ – 6i – 30i³ = x + iy

∴ 12(1) + 48i – 1 + 20( – i) – 6i – 30( – i) = x + iy  `...[(because "i"^2 = - 1),(therefore  "i"^3 = "i"^2 xx "i" = - "i"","  "i"^4 = ("i"^2)^2 = ( -1)^2= 1)]`

∴ 12 + 48i – 1 – 20i – 6i + 30i = x + iy

∴ 11 + 52i = x + iy

∴ x = 11, y = 52