# Homogeneous system of linear equations over $\mathbb{C}$

I have two systems of linear equations and I need to verify if they are indeed the same system, and if they are I must rewrite each equation as a linear combination of the others.

A: $\begin{array}{cc} 2x_1 + (-1+i)x_2 +x_4 = 0 \\ 3x_2 -2ix_3+5x_4=0\end{array}$

B: $\begin{array}{cc} (1+i/2)x_1 +8x_2 -ix_3 -x_4=0 \\ \frac{2}{3}x_1 -\frac{1}{2}x_2 + x_3 +7x_4=0 \end{array}$

My confusion lies in the fact that when I solve for $x_1,x_3$ in system A, I can get solutions in terms of $x_2,x_4$, whereas if I solve for $x_1,x_3$ in system B, $x_1$ is a function of $x_3$ and vice versa. This leads me to believe that the systems are not the same, but when I try and solve this system using mathematica, I am told that only trivial solution exists for both systems.

I was able to do this for the simpler examples with real variables but this one has me stumped.

Please note : this is the very beginning of my course so please refrain from using more advanced linear algebra techniques.

Thanks for the help!

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Does this mean you haven't done row reduction of matrices yet? – Gerry Myerson Jan 21 '13 at 23:10
Yes. Basically all i have is the theorem that says systems are equivalent if they have the same solutions. – MSEoris Jan 21 '13 at 23:11
By the way, you must be mis-using Mathematica, as a homogeneous linear system with more unknowns than equations is guaranteed to have non-trivial solutions. – Gerry Myerson Jan 21 '13 at 23:12
Thats what my intuition told me, but it seemed a pretty straightforward implementation. mm = {{2, (-1 + I), 0, 1}, {0, 3, -2 I, 5}} ; LinearSolve[mm, {0,0}]. – MSEoris Jan 21 '13 at 23:13
Your intuition is right. Take, for example, $x_2=17$, $x_4=42$, and you can solve A for $x_1$ and $x_3$. I don't know enough about Mathematica to tell you what's gone wrong. – Gerry Myerson Jan 21 '13 at 23:18

## 1 Answer

In B, multiply 2nd equation by $i$, add to 1st equation (so $x_3$ disappears), solve for $x_1$ in terms of $x_2$ and $x_4$, substitute this into either original equation of B, solve for $x_3$ in terms of $x_2$ and $x_4$, compare with your answer for A.

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