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

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)