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Solve this linear system over the field

crypt50

New member
Jun 29, 2013
21
\(\displaystyle \mathbb C: (1+i)x + 2y = 3, 3x + (1i)y = 2i\)

I don't know how to go about this, am I supposed to do addition of complex number? Please help.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
One way to proceed would be to multiply the second equation by $2i$, and then add the two equations:

\(\displaystyle (1+i)x+2y=3\)

\(\displaystyle 6ix-2y=-4\)

Then you will have eliminated $y$ and can solve for $x$, after which you may substitute for $x$ into either equation to determine $y$.
 

crypt50

New member
Jun 29, 2013
21
One way to proceed would be to multiply the second equation by $2i$, and then add the two equations:

\(\displaystyle (1+i)x+2y=3\)

\(\displaystyle 6i-2y=-4\)

Then you will have eliminated $y$ and can solve for $x$, after which you may substitute for $x$ into either equation to determine $y$.
I am not supposed to use Gaussian elimination.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I thought Gaussian elimination involved an augmented matrix. If you are not to use any type of elimination then use substitution instead.
 

eddybob123

Active member
Aug 18, 2013
76
Since this has two equations and two variables, it's easy to apply Cramer's Rule, since the determinant of a 2x2 matrix $\begin{bmatrix}a&b\\c&d \end{bmatrix}$ is just $ad-bc$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I am not supposed to use Gaussian elimination.
It would have helped if you had told us that!

In any case "complex numbers", as far as algebra is concerned, are just numbers. Any method that you could use with real coefficients works with complex coefficients.

If you are not allowed to use the easiest method (Gaussian elimination) you could use Cramer's rule, as eddybob123 suggested:
[tex]x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]
[tex]y= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]

(Editted thanks to eddybob123.)
 
Last edited:

eddybob123

Active member
Aug 18, 2013
76
I'm sure you meant y on the second equation!
 

crypt50

New member
Jun 29, 2013
21
It would have helped if you had told us that!

In any case "complex numbers", as far as algebra is concerned, are just numbers. Any method that you could use with real coefficients works with complex coefficients.

If you are not allowed to use the easiest method (Gaussian elimination) you could use Cramer's rule, as eddybob123 suggested:
[tex]x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]
[tex]x= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]
I thought I could figure it out. What if I don't want to use Cramer's rule.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle (1+i)x+2y=3\)

\(\displaystyle 6ix-2y=-4\)

Solve one of these equations for $2y$ and then substitute that into the other equation and solve for $x$.
 

VeryWell

New member
Aug 5, 2014
2