Another complex system of linear equations

  • #1
I guess i didn't get the last one completely because i've been having a hard time with this one. Solve

3x + iy + (2+i)z = 3i

-ix + y + z = 1

x + y + (2+i)z = i


i've tried dividing through by the leading coefficent, and re-arranging it...but can't seem to get the right answer! Thanks again for any help you can provide.
 

Answers and Replies

  • #2
cronxeh
Gold Member
961
10
[tex]
\left(
\begin{array}{xyzk}
3 & i & 2+i & 3i\\
-i & 1 & 1 & 1\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
3 & i & 2+i & 3i\\
1 & i & i & i\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex]
Multiply Row 2 by i, remember you can multiply by any constant - and i is just [tex]\sqrt{-1}[/tex]. i*i = -1, i*i*i = -i, i*i*i*i = 1

[tex]
\ = \left(
\begin{array}{xyzk}
2 & 0 & 2 & 2i\\
1 & i & i & i\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
2 & 0 & 2 & 2i\\
0 & i-1 & -2 & i-1\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & i-1 & -2 & i-1\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex]
R1 = R1 - R2 and R2 = R2 - R3 and R1 = 1/2 R1

[tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
-1 & i-1 & -3 & -1\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & i & i-1 & 0\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & -1 & -1-i & 0\\
1 & 1 & 2+i & 1
\end{array}
\right)
[/tex]
R2 = R2 - R1 and R2 = R2 + R3 and R2 = i*R2

[tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & -1 & -1-i & 0\\
0 & 1 & 1+i & 1-i
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 0 & 0 & 1-i\\
0 & 1 & 1+i & 1-i
\end{array}
\right)
[/tex]
R3 = R3 - R1 and R2 = R2 + R3

[tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 0 & 0 & 1-i\\
0 & 1 & 1+i & 1-i
\end{array}
\right)
[/tex]
You swap Row2 and Row3 and you get:

[tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 1 & 1+i & 1-i\\
0 & 0 & 0 & 1-i
\end{array}
\right)
[/tex][tex]
\ = \left(
\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 1 & 1+i & 0\\
0 & 0 & 0 & 0
\end{array}
\right)
[/tex]
R2 = R2 - R3 and you can eliminate Col 4 of R3

And viola! You end up with:
x + z = i
y + (1+i)z = 0
 
  • #3
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Arithmetic errors tend to be common on these types of problems (not just by students!!). The added complexity (haha!) of complex arithmetic only makes the problem worse.

I've found that the only reliable method (other than a symbolic calculator!) is to do it, check the answer, and if it's not right, try again. :frown:
 
  • #4
wow! thanks for the detailed solution! I need to practice these complex systems. Thanks!
 
  • #5
here's what i got...

[tex]\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R2 = R2 * i
[tex]\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\1 & i & i & i\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R1 = R1 - R2
[tex]\left(\begin{array}{xyzk}2 & 0 & 2 & 2i\\1 & i & i & i\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R2 = R2 - R3 and R1 = R1 / (2)
[tex]\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & i-1\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R2 = R2 - R1
[tex]\left(\begin{array}{xyzk}1 & 0 & 1 & i\\-1 & i-1 & -3 & -1\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R2 = R2 + R1
[tex]\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\1 & 1 & 2+i & 1\end{array}\right)[/tex]

R3 = R3 - R1
[tex]\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\0 & 1 & 1+i & 1-i\end{array}\right)[/tex]

R3 = R3 - [(-i-1)/2]R2
[tex]\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\0 & 0 & 0 & -i\end{array}\right)[/tex]

Doesn't that mean there is no solution?
 
  • #6
ah, the question was written down wrong in the matrix....it's suppose to be

[tex]\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & i\end{array}\right)[/tex]


(notice the last element is an 'i' not a 1) Thanks for everyones reply...i'm going to work with this, and see where it gets me...
 
  • #7
cronxeh
Gold Member
961
10
oh wow I'm sorry my mistake. Give me a minute to correct this
 
  • #8
cronxeh
Gold Member
961
10
[tex]\left(\begin{array}{xyzk}
3 & i & 2+i & 3i\\
-i & 1 & 1 & 1\\
1 & 1 & 2+i & i
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
3 & i & 2+i & 3i\\
1 & i & i & i\\
1 & 1 & 2+i & i
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
2 & 0 & 2 & 2i\\
1 & i & i & i\\
1 & 1 & 2+i & i
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & i-1 & -2 & 0\\
1 & 1 & 2+i & i
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & i-1 & -2 & 0\\
0 & 1 & 1+i & 0
\end{array}\right)[/tex]

[tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & i & -1+i & 0\\
0 & 1 & 1+i & 0
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & -1 & -i-1 & 0\\
0 & 1 & 1+i & 0
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 0 & 0 & 0\\
0 & 1 & 1+i & 0
\end{array}\right)[/tex][tex]\left(\begin{array}{xyzk}
1 & 0 & 1 & i\\
0 & 1 & 1+i & 0\\
0 & 0 & 0 & 0
\end{array}\right)[/tex]

x + z = i -> x = i - z
y + (1+i)z = 0 -> y = -z - iz

Check (from original equation):
3x + iy + (2+i)z = 3i -> 3(i-z) + i(-z -iz) + (2+i)z = 3i -> 3i - 3z -iz +1z + 2z +iz = 3i -> 3i = 3i

-ix + y + z = 1 -> -i(i-z) + (-z-iz) + z = 1 -> 1+iz -z -iz + z = 1 -> 1 = 1

x + y + (2+i)z = i -> i-z -z -iz +(2+i)z = i -> i-2z -iz + 2z +iz = i -> i=i
 
  • #9
thank you cronxeh, you've been very helpful!
 

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