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8 January, 03:08

Find all solutions for the following systems with complex coefficients. (If there are an infinite number of solutions, use t as your parameter.) ix1 + x2 = 4 2x1 + (1 - i) x2 = 3i

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  1. 8 January, 04:06
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    The solutions are:

    x1 = (-1/2) (5 + 3i)

    and

    x2 = (1/2) (11 + 5i)

    Step-by-step explanation:

    Given the equations:

    ix1 + x2 = 4 ... (1)

    2x1 + (1 - i) x2 = 3i ... (2)

    From (1), make x2 the subject, to have:

    x2 = 4 - ix1 ... (3)

    Using (3) in (2)

    2x1 + (1 - i) (4 - ix1) = 3i

    2x1 + 4 - ix1 - 4i + i²x1 = 3i

    2x1 + 4 - ix1 - 4i - x1 = 3i

    (because i² = - 1)

    (2 - i - 1) x1 + 4 (1 - i) = 3i

    (1 - i) x1 + 4 (1 - i) = 3i

    Divide both sides by (1 - i)

    x1 + 4 = 3i / (1 - i)

    Multiply both the numerator and denominator of the right hand side by the conjugate of (1 - i). The conjugate of (1 - i) is (1 + i), the aim of this multiplication is to makes the denominator a real number, rather than complex.

    x1 + 4 = 3i (1 + i) / (1 - i) (1 + i)

    x1 + 4 = (3i - 3) / 2

    x1 = (3/2) (i - 1) - 4

    = [3 (1 - i) - 8]/2

    = (3 - 3i - 8) / 2

    = (-5 - 3i) / 2

    x1 = (-1/2) (5 + 3i) ... (4)

    Using this in (3)

    x2 = 4 - ix1

    x2 = 4 - i (-1/2) (5 + 3i)

    = 4 + (1/2) (5i - 3)

    = (8 + 5i + 3) / 2

    = (11 + 5i) / 2

    x2 = (1/2) (11 + 5i)
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