Another complex system of linear equations

In summary, the conversation involves solving a system of complex equations using row operations. The process involves multiplying rows by constants, adding and subtracting rows, and rearranging terms. The final solution is x = i-z, y = -z-iz, and z can be any complex number. The conversation also includes a correction to the original equation, which results in a solution where x = i-z, y = -z-iz, and z = i.
  • #1
3.14159265358979
43
0
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.
 
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  • #2
[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
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
oh wow I'm sorry my mistake. Give me a minute to correct this
 
  • #8
[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!
 

1. What is a complex system of linear equations?

A complex system of linear equations is a set of equations with multiple variables and coefficients that are related to each other through linear relationships. This means that each equation can be written in the form of y = mx + b, where x is a variable, m is a coefficient, and b is a constant.

2. What is the purpose of solving a complex system of linear equations?

The purpose of solving a complex system of linear equations is to find the values of the variables that satisfy all of the equations in the system. This allows us to determine the relationship between the variables and make predictions or solve problems related to the system.

3. How do you solve a complex system of linear equations?

There are multiple methods for solving a complex system of linear equations, including substitution, elimination, and graphing. These methods involve manipulating the equations in the system to eliminate variables and eventually find a solution that satisfies all of the equations.

4. What types of problems can be solved using complex systems of linear equations?

Complex systems of linear equations can be used to solve a variety of real-world problems, such as determining the optimal production levels for a business, finding the intersection point of two moving objects, or calculating the cost of a meal at a restaurant.

5. What are some common challenges when solving a complex system of linear equations?

Some challenges when solving a complex system of linear equations include having multiple variables and equations to keep track of, making sure the equations are accurately set up, and avoiding errors during the solving process. It is also important to check the solutions at the end to ensure they satisfy all of the equations in the system.

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