Complex number equation: Gauß

ozfingwe · #1 $Old$ November 1st, 2009, 08:47 AM

hi there,

could anybody please give me a hint on how to go about solving the following system of complex number equations using Gauß' elimination algorithm (though I guess Cramer works too)?

x + i * y + (2-i) * z = 2 + i
(1+i) * x + 3i * z = 1 + 4i
-2x + (1-i) * y + (1+2i) * z = i

I got stuck after adding twice the first to the third and swapping third and second thereafter... :-/

thank you -
oz.

tonio · #2 $Old$ November 1st, 2009, 12:04 PM

Quote:

Originally Posted by ozfingwe $View Post$

hi there,

could anybody please give me a hint on how to go about solving the following system of complex number equations using Gauß' elimination algorithm (though I guess Cramer works too)?

x + i * y + (2-i) * z = 2 + i
(1+i) * x + 3i * z = 1 + 4i
-2x + (1-i) * y + (1+2i) * z = i

I got stuck after adding twice the first to the third and swapping third and second thereafter... :-/

thank you -
oz.

Write down your augmented matrix:

$\left(\begin{array}{cccc}1&i&2-i&2+i\\1+i&0&3i&1+4i\\\!\!-2&1-i&1+2i&i\end{array}\right)$

Now, substract from Row 2 (1+i)*Row 1 and add twice Row 1 to Row 3:

$\left(\begin{array}{cccc}1&i&\;2-i&\!\!2+i\\0&1-i&\!-3+2i&i\\0&1+i&5&4+3i\end{array}\right)$

Finally, add -i*Row 2 to Row 3:

$\left(\begin{array}{cccc}1&i&\;2-i&\!\!2+i\\0&1-i&\!-3+2i&i\\0&0&7+3i&5+3i\end{array}\right)$

Now solve beginning from Row 3 and backwards: $(7+3i)z=5+31\Longrightarrow z=\frac{5+3i}{7+3i}=\frac{-22+3i}{29}$ , now substitute this in Row 2 and find y, and then substitute in Row 3 and find x.

Tonio