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I need to show that two systems of linear equations are equivalent, however this is over the complex field.

How would I solve this?

One of the systems is:

$$\left\{\begin{align*} &2x_1+(-1+i)x_2+x_4=0\\ &3x_2-3ix_3+5x_4=0\;. \end{align*}\right.$$

Yet no matter how many ways I look at it, I can't find a way to reduce it enough in the matrix that it's practical to solve. Is there another way of showing that two systems are equivalent?

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You can bring this system into reduced row echelon form and do the same for your other system. – Michael Joyce Jan 13 '13 at 5:56

You solve it in the exact same way as in the real case. That is, since it's underdetermined, you need to pick two variables you'll leave undetermined (let's say $x_3$ and $x_4$), and form the matrix equation

$$\left[\begin{array}{cc}2 & -1+i \\0 & 3\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}-x_4\\3ix_3-5x_4\end{array}\right]$$

and then invert the matrix to get

$$\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\frac{1}{6}\left[\begin{array}{cc}3 & 1-i \\0 & 2\end{array}\right]\left[\begin{array}{c}-x_4\\3ix_3-5x_4\end{array}\right] = \frac{1}{6}\left[\begin{array}{c}(-3+3i)x_3+(-8+5i)x_4\\6ix_3-10x_4\end{array}\right]$$

Now you can do the same thing to your other system and verify that you get the same expressions for $(x_1,x_2)$.

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Hmm, we haven't learned that technique yet. Would it be possible to put both systems into the same matrix so that I have 4 equations and then solve for the variables? – Duncan Forster Jan 13 '13 at 18:50
No, you won't be able to, because at least two of your rows will be linearly independent. You could put both $2\times 4$ systems in row-echelon form and compare them (Michael's suggestion). – user7530 Jan 13 '13 at 21:13

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