Solving a system of linear equations with complex number? In my circuit analysis class I consistently need to solve system of complex equations, and I can't use MATLAB or anything for it. Suppose I have the following system: (Note that $j = \sqrt{-1})$
$Va,Vs,Vo$ are unknowns in the form of $a + jb$
$$\frac{Va-Vs}{-j15} + \frac{Va}{33} + \frac{Va-Vo}{-j25}=0$$
$$\frac{Vo-Va}{-j25} + \frac{Vo-Vs}{10} = 0$$
Exactly how am I supposed to solve this equation if there are complex numbers in it (by hand)?
 A: The first task is to clear the fractions, making it easier to collect like terms.  Note that dividing by $-j$ amounts to multiplying by $j$, and that integer denominators can be cleared by multiplying by their least common multiple.  Thus:
$$ \frac{V_a - V_s}{-15j} + \frac{V_a}{33} + \frac{V_a - V_o}{-25j} = 0 $$
becomes, after multiplying by $33\cdot 25$:
$$ (55j)(V_a - V_s) + 25V_a + (33j)(V_a - V_o) = 0 $$
After collecting like terms:
$$ (25+88j)V_a - 33j\; V_o - 55j\; V_s = 0 \tag{1} $$
Similar treatment of the second equation:
$$ \frac{V_o - V_a}{-25j} + \frac{V_o - V_s}{10} = 0 $$
results in:
$$ (2j)(V_o - V_a) + 5(V_o - V_s) = 0 $$
and then:
$$ -2j\; V_a + (5+2j)V_o - 5V_s = 0 \tag{2} $$
A moment's study reveals that (1),(2) form an independent pair of homogeneous linear equations in the three unknowns $V_a,V_o,V_s$.  Thus solutions will collectively be all scalar multiples of some particular solution, i.e. a one parameter family of solutions.
One approach would be to set (for example) $V_s = 1$ and solve for the other two values.  This yields a linear system of two equations in two unknowns:
$$ (25+88j)V_a - 33j\; V_o = 55j \tag{1'} $$
$$ -2j\; V_a + (5+2j)V_o = 5 \tag{2'} $$
The augmented matrix for this system is:
$$ \begin{bmatrix} 25+88j & -33j & \mid & 55j \\
                     -2j  & 5+2j & \mid & 5 \end{bmatrix} $$
Putting the augmented matrix in reduced row echelon form is much the same use of elementary row operations, modified only insofar as complex arithmetic replaces real arithmetic.
Since it is desirable to do this "by hand", we will introduce a leading one in the first row by adding $44 - 6j$ times the second row to the first row (rather than by dividing the first row by $25 + 88j$):
$$ \begin{bmatrix}     1 & 244-5j & \mid & 220-5j \\
                     -2j &  5+2j  & \mid & 5 \end{bmatrix} $$
Now I would add $2j$ times the first row to the second, to eliminate the entry under the new leading one:
$$ \begin{bmatrix}     1 & 244-5j  & \mid & 220-5j \\
                       0 & 15+490j & \mid & 15+440j \end{bmatrix} $$
Unfortunately we cannot put off doing a complex division much longer.  To get a leading one in the second, I would divide by the coefficient $15+490j$:
$$ \frac{15+440j}{15+490j} = \frac{3+88j}{3+98j} = 1 - \frac{10j}{3+98j} $$
The denominator of this last fraction can be rationalized:
$$ \frac{10j}{3+98j} = \frac{10j(3-98j)}{3^2 + 98^2} = \frac{980+30j}{9613} $$
The (unreduced) row echelon form of the matrix is:
$$ \begin{bmatrix}     1 & 244-5j  & \mid & 220-5j \\
                       0 &    1    & \mid & 1 - \frac{980+30j}{9613}
 \end{bmatrix} $$
