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Question
Find x and y if `|(4i, i^3, 2i),(1, 3i^2, 4),(5, -3, i)|` = x + iy where i2 = –1
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Solution
`|(4i, i^3, 2i),(1, 3i^2, 4),(5, -3, i)|`
`=|(4i, i^3, 2i),(1, 3i^2, 4),(5, -3, i)|` ...[∵ i2 = −1]
`= |(4i,-i,2i),(1,-3,4),(5,-3,i)|`
`= 4i|(-3, 4),(-3, i)|-(-i)|(1, 4),(5, i)| + 2i|(1, -3),(5, -3)|`
`= |(-3,4),(-3,i)| = (-3)(i)-(4)(-3)=-3i+12`
4i ⋅ (−3i + 12) = 4i ⋅ 12 − 4i ⋅ 3i = 48i − 12i2 = 48i + 12
`|(1,4),(5,i)| = (1)(i) - (4)(5) = i-20`
+i ⋅ (i − 20) = i2 − 20i = −1 − 20i
`|(1,-3),(5,-3)| = (1)(-3) - (-3)(5) = -3+15 = 12`
2i ⋅ 12 = 24i
(48i + 12) + (−1 − 20i) + 24i = (48i − 20i + 24i) + (12 − 1) = 52i + 11
x = 11, y = 52
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